Thermal Properties Of Matter - Class 11 Medical Physics - Extra Questions

Who is telling the truth? The temperature of a substance is measured in joules.

The sentence is $$false$$ . The heat energy of a substance is measured in joules.


At what temperature is the Fahrenheit scale reading equal to 
twice that of the Celsius scale.

Let the reading on the Celsius scale be $$x$$ and the reading on the Fahrenheit scale be $$y$$. Then $$y=\dfrac 95x+32$$. If we required $$y=3x$$, then we have
$$2x=\dfrac 95 x+32\ \Rightarrow \ x=(5)\ (32)=160^oC$$
which yields $$y=2x=320^oF$$


At what temperature is the Fahrenheit scale reading equal to 
half that of the Celsius scale?

In this case,
 we required $$y=\dfrac 12 x$$ and find
$$\dfrac 12x=\dfrac 95x+32\ \Rightarrow \ x=-\dfrac {(10)(32)}{13}\cong -24.6^oC$$
which yields $$y=x/2=-12.3^oF$$.


A container with $$1 kg$$ of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is $$700 Wm^{-2}$$ and it is absorbed by the water over an effective area of $$0.05 m^2$$. Assuming that the heat loss from the water to the surroundings is governed by Newtons law of cooling, the difference (in $$^{\circ}C$$) in the temperature of water and the surroundings after a long time will be ______. (Ignore effect of the container, and take constant for Newtons law of cooling $$= 0.001 s^{-1}$$, Heat capacity of water $$= 4200 J \ kg^{-1} K^{-1}$$)

$$\dfrac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$$

For small temperature change.

$$\dfrac{dQ}{dt} = e \sigma AT^3 \Delta T$$ ___(i)

$$\dfrac{mcdT}{dt} = e \sigma AT^3 \Delta T$$

$$\dfrac{dT}{dt} = \dfrac{e \sigma A T^3}{mC} \Delta T$$

$$\dfrac{e \sigma AT^3}{mC} \rightarrow$$ constant for Newton law of cooling

$$\dfrac{e\sigma AT^3}{mC} = 0.001$$

$$e \sigma AT^3 = mC \times 0.001$$

$$ = 1 \times 4200 \times 0.001$$

$$e \sigma AT^3 = 4.2$$ ___(ii)

$$\dfrac{dQ}{dt} = 700 \times 0.05 = 35 Watts $$ ___(iii)

Put the value of equation (ii), and equation (iii)
in equation (i)

$$35 = 4.2 \Delta T$$

$$\dfrac{35}{4.2} = \Delta T$$

$$\Delta T = 8.33$$


Express $$281 K$$ in $$^oC$$

$$K=C+273$$
$$281=C+273$$
$$C= 281 - 273 = 8^oC$$


Express $$0C$$ in kelvin.

$$K=C+273$$
at $$C=0, K=273$$


The body temperature of a healthy person is $$98.6F$$. Calculate the corresponding temperature on the Kelvin scale.

$$K = (F - 32) \times \dfrac{5}{ 9} + 273.15$$
putting $$F=98.6$$ we get $$K=310$$


Express $$45^oC$$ in kelvin.

$$K=C+273 $$
at $$C = 45$$, 
$$K=318$$


How can the radiating power of a surface be increased?

   It is found experimentally (e.g. Leslie's experiment) that a rough and dark surface is a good emitter and absorber of heat radiation . Therefore by making the surface rough and black we can increase the radiating power of surface .


What is the value of $$0^{\circ}$$F in celsius scale ?

$$\dfrac{F-32}{9} =\dfrac{C}{5}$$
$$ \Rightarrow \frac{0-32}{9} =\dfrac{C}{5}$$
$$ \Rightarrow C =-\dfrac{5}{9}32=-17.8^{\circ}C$$


The fundamental intervals of two thermometers X and Y are $$70^{\circ}$$ and $$140^{\circ}$$ respectively. Their ice points are $$10^{\circ}$$ and $$0^{\circ}$$ respectively. If Y reads $$90^{\circ}$$. What would X read?

As given 
$$\dfrac{X-10}{70}=\dfrac{Y-0}{140}$$
Now when $$Y=90^o$$
So $$\dfrac{X-10}{70}=\dfrac{90}{140}$$
$$X=55^0$$


A difference of $$1^{\circ}C$$ is _____ as the difference of $$1\ K$$.

We know that $$K=273+^oC$$
So if we will increase temperature by $$1^oC$$ then in Kelvin temperature will also increase by 1.
For example let say temperate is $$^oC=27$$
So in Kelvin $$T=27+273=300K$$
Now if we increase the temperature by $$1^oC$$, that is $$27+1=$$ $$28^oC$$
Now in Kelvin, $$T_1=28+273=301K$$
So in Kelvin, the increase in temperature is also 1.


Convert each of the following temperature in $$^{\circ}F$$ to the celsius and kelvin scale $$68^{\circ}F, 5^{\circ}F, 176^{\circ}F$$

We know that , $$9C=5(F-32)$$
Where C is temperature in Celsius and F is in Fahrenheit 
Now when Temperature is  $$T_F=68^oF$$
And Let T be the temperature in Celsius scale
$$9T=5(68-32)$$ 
So $$T=20^oC$$
Now when Temperature is  $$T_F=5^oF$$
And Let T be the temperature in Celsius scale
$$9T=5(5-32)$$ 
So $$T=-15^oC$$
Now when Temperature is  $$T_F=176^o$$
And Let T be the temperature in Celsius scale
$$9T=5(176-32)$$ 
So $$C=80^oC$$


Triple point of water is

The triple point of water is $$273.16K$$.


Fill in the blanks:
Heat is a form of ________.

Heat is a form of energy.


In which mode  of heat transfer , no medium is required ?

Radiation


Observe the picture given in above Figure. Water is being boiled in a pan of wide base.
i) Which position $$P$$ or $$T$$ will feel warmer?
ii) Fill up the boxes $$P$$ and $$T$$ to indicate the mode of flow of heat to the hand.
1788025_52e51eeffa8042daade929a235864ee6.png

According to the given condition:
i) Between position $$P$$ or $$T$$, position $$P$$ will be warmer due to the hot air rising up.
ii) P= Convection and T= Radiation.


The fundamental intervals of two thermometers X and Y are $$70^o$$ and $$140^o$$ respectively. Their ice points are $$10^o$$ and $$0^o$$ respectively, If Y reads $$90^o$$. What would X read?

Given Fundamental interval of
$$X=UFP_1-LFP_1=70^o$$
Fundamental interval of
$$Y=UFP_2=LPF_2=140$$
LFP of $$X=10^o$$
LFP of $$Y=0^o$$
Reading on $$Y=90^o$$

$$\dfrac {X-LPF_1}{UFP_1-LFP_1}=\dfrac {Y-LFP_2}{UFP_2-LFP_2}$$

$$\dfrac {X-10}{70}=\dfrac {90-0}{140}$$

$$X-10=\dfrac {90\times 70}{140}$$

$$X=45+10=55^o$$


Calculate the temperature on Celsius scale whose value is half the absolute reading?

Basic conversion formula: $$K=C+273.15$$
Given $$C=\dfrac K2=\dfrac {C+273.15}2$$ 
$$\implies C=273.15°C$$, is the required temerature on Celsius scale


A faulty thermometer has its fixed points marked as $$10^o$$ and $$90^o$$. The temperature of a body as measured by the faulty thermometer si $$61^o$$. Find the correct temperature of the body on celsius scale.

Let $$\theta_0$$ be the lower fixed point of the faulty thermometer and 'n' be the number of divisions between its lower and the upper fixed points. If a temperature C on celsius scale corresponds to temperature $$\theta$$ on the scale of faulty thermometer, then
$$\dfrac {C}{100}=\dfrac {\theta-\theta_0}{n}$$

Here, $$\theta_0=10^o, n=90-10=80$$ and $$\theta=61^o$$
Therefore, the equation becomes
$$\dfrac {C}{100}=\dfrac {61-10}{90-10}$$

$$C=\dfrac {56}{80}\times 100$$

$$C=63.75^o C $$


The temperature of a body changes from $$15^oC$$ to $$75^oC$$. What is the corresponding change in (a) Fahrenheit scale (b) kelvin scale?

$$\Delta C=75-15=60^oC$$

(a) $$\dfrac {60}{100}=\dfrac {F-32}{180}$$

$$\Delta F=140^oF$$

(b) $$\Delta K=\Delta C=60 K$$


At what temperature in degree Kelvin the Fahrenheit and Kelvin scales of temperature give the same reading ?

$$F=1.8\  ^oC-32$$  and $$K=\ ^oC+273$$
For same reading,  $$F=K \Rightarrow 1.8\  ^oC-32= \ ^oC+273,    ^oC=301.25 \sim 301$$
Temperature in degree Kelvin $$=301+273=574$$


Describe an experiment to show that a blackened surface is a better emitter of heat than a polished one.

Black body is the best emitter because if something is black, it means it's strongly coupled to electromagnetic field. A body which a good absorbed of heat is also a good emitter because absorbing is just opposite process of emitting. And also by Kirchoff's law of radiation, make it evident that a black body is not only a perfect absorbed but also a perfect emitter of heat.


At what temperature the Fahrenheit and Celsius scale of temperature gives the same reading ?

Since the temperature scales are linear,

$$\dfrac { C-0 }{ 100-0 } =\dfrac { F-32 }{ 212-32 } $$

$$F=\dfrac { 9 }{ 5 } C+32$$

For reading to be same on both the scales,

$$C=\dfrac { 9 }{ 5 } C+32$$

$$C=-{ 40 }^{ 0 }C$$


At what temperature is the Fahrenheit scale reading equal to twice the reading of Celsius scale ?

$$\dfrac{F-32}{9} =\dfrac{C}{5}$$
$$ \Rightarrow \dfrac{2C-32}{9} =\dfrac{C}{5}$$
$$\dfrac{C}{5} =\dfrac{32}{9}$$
$$\Rightarrow C = \dfrac{160}{9} = 17.7^{\circ}$$


Change each of the given temperature to the Celsius and Kelvin scales: $$68^{\circ} F, 5^{\circ} F$$ and $$176^{\circ} F.$$

Following are the expressions for the required conversions.

$${ T }_{ C }=\dfrac { 5 }{ 9 } ({ T }_{ F }-32)$$

$${ T }_{ K }=\dfrac { 5 }{ 9 } ({ T }_{ F }+459.67)$$

Thus, putting in values,

$${ T }_{ C1 }=\dfrac { 5 }{ 9 }(68-32)={ 20 }^{ o }C$$

$${ T }_{ C2 }=\dfrac{ 5 }{ 9 }(5-32)={ -15 }^{ o }C$$

$${ T }_{ C3 }=\frac{ 5 }{ 9 }(176-32)={ 80 }^{ o }C$$

$${ T }_{ K1 }=\dfrac { 5 }{ 9 } (68+459.67)=293K$$

$${ T }_{ K2 }=\dfrac { 5 }{ 9 } (5+459.67)=258K$$

$${ T }_{ K3 }=\dfrac { 5 }{ 9 } (176+459.67)=353K$$


Draw a clear labelled diagram of an experiment to compare the absorbing powers of two different surfaces, a shiny white and a dull black.


1154985_202979_ans_8e8ba6bae6ab426888b044c266aa5863.png


Describe an experiment to show that a black (or dull) surface is a better absorber of heat radiations than a polished one.

When heat radiation fall on polished surface or white objects, most of the heat energy is reflected black but when these heat energy are absorbed and very less amount of energy is reflect back apart from this black body absorbs more heat because black colours absorbs all colours of light energy, the more light is absorbed, the more heat will be absorbed.


A Faulty thermometer reads $$5^{\circ}C$$ in melting ice and $$99^{\circ}C$$ in steam. Find the correct temperature in $$^{\circ}F$$ when this faulty thermometer reads $$52^{\circ}C$$.

Since the temperature scale is linear,

$$\dfrac { T-5 }{ 99-5 } =\dfrac { F-32 }{ 212-32 } $$

$$F=\dfrac { 180 }{ 94 } (T-5)+32$$

$$F=\dfrac { 180 }{ 94 } (52-5)+32={ 122 }^{ 0 }F$$

Answer is $${ 122 }^{ 0 }F$$


At what temperature do the Celsius and Fahrenheit reading have the same numerical value ?

From the relation 

$${ T }_{ F }=\dfrac { 9 }{ 5 } { T }_{ C }+32$$

For the two to be equal,

$${ T }_{ F }=\dfrac { 9 }{ 5 } { T }_{ F }+32$$

$$-\dfrac { 4 }{ 5 } { T }_{ F }=32$$

$${ T }_{ F }={ -40 }^{ o }F={ T }_{ C }={ -40 }^{ o }C$$

Answer is $${ -40 }^{ o }C$$


The efficiency of an ideal engine is $$75\%$$ and it emits heat $$2 \times 103\ W$$ to sink at $$283\ K$$, then calculate :
Temperature of source.

GIven, $$\eta =75\%=\dfrac{75}{100}=\dfrac{3}{4};\ T_2=283\ K;$$
$$Q_2=2 \times 10^3\ W$$ or $$J.s^{-1}$$
$$(i)\ \because \eta=1-\dfrac{T_2}{T_1}\ \ \Rightarrow \dfrac{T_2}{T_1}=1- \eta$$
$$\therefore \dfrac{283}{T_1}=1-\dfrac{3}{4}=\dfrac{1}{4}$$
or $$T_1=4 \times 283=1135\ K$$


The steam point and the ice point of a mercury thermometer are marked as $$80^{\circ}$$ and $$20^{\circ}$$. What will be the temperature in centigrade mercury scale when this thermometer reads $$32^{\circ}$$

Explanation for correct option:

$$\textbf{Step1: Find temperature in centigrade mercury scale }$$

By the standard temperature to a temperature measured by mercury formula, which is $${T_c} = \dfrac{{T - {T_0}}}{{{T_{100}} - {T_0}}} \times 100$$

Where,

$${T_c}$$ is the temperature in centigrade

$$T$$ is temperature measured

$${T_0}$$ is ice point

$${T_{100}}$$ is steam point.

$$ \Rightarrow {T_c} = \dfrac{{32 - 20}}{{80 - 20}} \times 100 = 20^\circ C$$

The temperature in centigrade mercury scale when this thermometer reads $$32^\circ C$$ will be $$20^\circ C$$.



Change each of the given temperature to the Fahrenheit and Rankine scales: $$30^{\circ} C, 5^{\circ} C$$ and $$-20^{\circ} C.$$

Following are the relations for required conversions.

$${ T }_{ F }=\dfrac { 9 }{ 5 } { T }_{ C }+32$$

$${ T }_{ R }=\dfrac { 9 }{ 5 } ({ T }_{ C }+273.15)$$

Thus, for given values,

$${ T }_{ F1 }=\dfrac { 9 }{ 5 } (30)+32={ 86 }^{ o }F$$

$${ T }_{ F2 }=\dfrac { 9 }{ 5 } (5)+32={ 41 }^{ o }F$$

$${ T }_{ F3 }=\dfrac { 9 }{ 5 } (-20)+32={ -4 }^{ o }F$$

$${ T }_{ R1 }=\dfrac { 9 }{ 5 } (30+273.15)={ 546 }^{ o }R$$

$${ T }_{ R2 }=\dfrac { 9 }{ 5 } (5+273.15)={ 501 }^{ o }R$$

$${ T }_{ R3 }=\dfrac { 9 }{ 5 } (-20+273.15)={ 456 }^{ o }R$$


Derive a relation between the Celsius and Fahrenheit Scales of temperature using the fixed points in the two different scales. 

$$ \displaystyle \frac{C-L.F.P}{U.F.P-L.F.P}=\frac{F-L.F.P}{U.F.P-L.F.P} $$ ; 
Where L.F.P = Lower Fixed Point 
            U.F.P = Upper Fixed Point 
$$ \displaystyle \frac{C-0}{100-0}=\frac{F-32}{212-32} $$
$$ \displaystyle \frac{C}{100}=\frac{F-32}{180} $$
$$ \displaystyle \frac{C}{5}=\frac{(F-32)}{9} $$


The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

The conversion formulae are 
$$t_{c}=t_{k}-273.15$$
$$t_{f}=\dfrac{9}{5}t_{c}+32$$
Substituting the given values, we get 
$$\text{Neon:}\ 24.57\ K=-248.43\ ^o C = -415.44\ ^o F$$
$$\text{CO:}\ 216.55\ K=-56.45\ ^o C=-69.8\ ^oF$$


Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between  $$ \displaystyle T_{A} $$ and  $$ \displaystyle T_{B} $$ ?

Given data,
triple point of water on scale $$A = 200A$$
triple point of water on scale $$B = 350B$$
we know, that triple point of water on two scales$$ A $$ and$$ B $$, then
$${\dfrac{273.16}{200}}TA = { \dfrac{273.16}{350}}TB$$
$$\dfrac{TA}{TB} = \dfrac{200}{350}$$
$$= \dfrac{20}{35}$$
$$=\dfrac{ 4}{7}$$

$$TA =\dfrac{ 4TB}{7}$$



A faulty thermometer has its fixed points marked as $$10^{\circ}$$ and $$90^{\circ}$$. The temperature of a body as measured by the faulty thermometer is $$61^{\circ}$$. Find the correct temperature of the body on Celsius scale.

Let the temperature on faulty thermometer be T and correct thermometer be C.
So $$\dfrac{T-10}{90-10}=\dfrac{C-0}{100-0}$$
So now the $$T=61^o$$
$$C=63.75^o$$, reading on correct thermometer. 


The temperature of a body changes from $$15^{\circ}C$$ to $$75^{\circ}C$$. What is the corresponding change in (a) Fahrenheit scale (b) Kelvin scale?

We know $$9C=5(F-32)$$
When $$C=15^0$$
$$9\times 15=5(F-32)$$
$$F=59^0$$
When $$C=75^0$$
$$9\times 75=5(F-32)$$
$$F=167^0$$
Change in Fahrenheit scale $$=167-59=108 ^o F$$
For Kelvin scale:
$$K=273+C$$
When C is $$15^o$$
$$K=273+15=288$$
When C is $$75^o$$
$$K=273+75=348$$
So change in kelvin scale is $$348-288=60K$$ 


The boiling point of turpentine on Fahrenheit scale is double its value on Celsius scale. If the boiling point of tar on Celsius scale is $$140^{\circ}$$ more than that of turpentine on same scale, find the boiling point of tar on Kelvin scale.

We know that $$9C=5(F-32)$$
Let the temperature in Celsius scale be T and in Fahrenheit be 2T.
So $$9T=5(2T-32)$$
$$T=160^oC$$ 
Now the boiling temperature of turpentine $$T_1=160+140=300^0C$$
We know for kelvin, $$K=273+C=273+300$$
In Kelvin scale it is  $$573K$$. 


Name the modes of transfer of heat in the following:
(a) Solid,
(b) Liquid,
(c) Gas,
(d) Vacuum.

Following are the different modes of transfer of heat with different mediums:


(a) Solid: 

  • In solid, heat can transfer only by conduction.

  • Conduction is the process of transferring heat from hot end to cold end from particle to particle of a medium.

  • Heat transfer by conduction mainly occurs in solid.


(b) Liquid:

  • In liquid, heat transfers mainly by convection.

  • Because convection is the process of heat transfer by actual movement of medium particles.

  • To some extent liquid also transfers heat by conduction when they are at static condition.


 

(c) Gas: 

  • In gases, heat transfers mainly by convection.

  • Because convection is the process of heat transfer by actual movement of medium particles.

  • To some extent gases also transfer heat by conduction when they are at static condition.



 

(d) Vacuum: 

  • In vacuum, heat can transfer only by radiation.

  • Because radiation is the process of heat transfer in which heat directly transfers from a hot body to another cold body without affecting the medium.

  • In radiation heat transfer no medium is required.






Who is telling the truth? Joule is the unit of heat.

Sentence is telling the $$truth$$ as joule is the unit of heat .


Explain conduction, convection and radiation with examples

  • Conduction transfers heat via direct molecular collision. An area of greater kinetic energy will transfer thermal energy to an area with lower kinetic energy. Higher-speed particles will collide with slower speed particles. The slower-speed particles will increase in kinetic energy as a result. Conduction is the most common form of heat transfer and occurs via physical contact. Examples would be to place hand against a window or to place metal into an open flame.

  • When a fluid, such as air or a liquid, is heated and then travels away from the source, it carries the thermal energy along. This type of heat transfer is called convection. The fluid above a hot surface expands, becomes less dense, and rises. At the molecular level, the molecules expand upon introduction of thermal energy. As temperature of the given fluid mass increases, the volume of the fluid must increase by same factor. This effect on the fluid causes displacement. As the immediate hot air rises, it pushes denser, colder air down. Example would be the flow of wind.

  • Thermal radiation generates from the emission of electromagnetic waves. These waves carry the energy away from the emitting object. Radiation occurs through a vacuum or any transparent medium (either solid or fluid). Thermal radiation is the direct result of random movements of atoms and molecules in matter. Movement of the charged protons and electrons results in the emission of electromagnetic radiation. All materials radiate thermal energy based on their temperature. The hotter an object, the more it will radiate. The sun is a clear example of heat radiation that transfers heat across the solar system. At normal room temperatures, objects radiate as infrared waves.




In a scientific book that describes a temperature scale called $$Z$$, boiling and freezing point of water are referred as $${65}^{o}Z$$ and $$-{15}^{o}Z$$ respectively. (a) To what temperature on Fahrenheit scale would a temperature $$-{95}^{o}Z$$ compound? (b) What temperature change on the $$Z$$ scale that would correspond to a change of $${40}^{o}$$ on Celsius scale?

Freezing Point $$=-15°z$$
Boiling Point $$=65°z$$
$$(a) \left [ T°F \right] =\left[ T°Re \right] \times \cfrac { 9 }{ 4 } +32$$
      $$\left[ T°F \right] =\left[ -95°z \right] \times \cfrac { 9 }{ 4 } +32$$
      $$\left[ T°F \right] =-181.75°F$$
      $$\therefore -95°z=-181.75°F$$


$$(b) \left[ T°Z \right] =\left[ T°C \right] \times \cfrac { 4 }{ 5 } $$
        $$T°Z=40°C\times\cfrac { 4 }{ 5 } $$
        $$T°Z=32°Z$$
        $$\therefore 40°C=32°Z$$



A Centigrade thermometer has its lower and upper fixed points marked $$-0.5^{\circ}C$$ and $$100.5^{\circ}$$. What is the true temperature when this thermometer reads $$30^{\circ}$$? The bore of the thermometer is uniform.

Ideally a Thermometer should be divided into 100 equal parts, the upper fixed point should be $$100^o$$, and the lower should be $$0^o$$
$$100.5-(-0.5)=101$$
This means that the given thermometer is divided into 101 equal parts.
The lower fixed point is $$-0.5^o$$. This means that the given thermometer is half step behind the ideal one.
Also note that any one step on this thermometer means $$\cfrac{100}{101}$$  step of the ideal thermometer.
Hence if it shows the reading as $$30^o$$ then it should mean $$(\cfrac{30\times100}{101})+(\cfrac{0.5\times100}{101})=30.198$$

Answer:-  30.198$$^o$$C



The steam point and the ice point of a mercury thermometer are marked as $$80^{\circ}$$ and $$10^{\circ}$$. At what temperature on centigrade scale the reading of this thermometer will be $$59^{\circ}$$?

Let the relation between the thermometer reading and centigrade be
$$y = ax + b$$
given at $$x = 100, y = 80$$ and at $$x = 0, y = 10$$
$$\therefore 80 = 100a + b, 10 = b\Rightarrow a = 0.7$$
Now, we have to find $$x$$ when $$y = 59$$
$$\therefore 59 = 0.7x + b\Rightarrow x = 70$$
$$\therefore$$ The answer is $$70^{\circ}C$$.


Compare the rates of emission of heat of a black body maintained at $$727^oC$$ with $$227^oC$$. If the black body is surrounded by an enclosure (black) at $$27^oC$$, what would be the ratio of their rates of loss of heat.


1961591_1226066_ans_3a91e94943064c28a9c437c6bb57bdfb.jpg


Explain the following temperature vs time graph.
1298395_c654c3fff1c8481282eac96c55d30834.PNG

With the increase in temperature,

when the time range is between points A and B, the material is in form of solid+liquid state,

when the time range is between points B and C, the material is in form of liquid+gaseous state,

when the time ranges after the point C, the material is in form of boiling water+vapour state,




Calculate the heat supplied in figure shown for complete cycle.
1361726_a6477026ae9945e39d6ebc87eb5b839d.jpg

$$dQ = du + dw$$
$$\therefore \triangle Q = \triangle w$$
$$\triangle Q =$$ Area under $$P - V$$ curve.
$$\triangle Q = -\dfrac {1}{2} \times base\times height$$
$$= -\dfrac {1}{2} \times 2\times 5\times 10^{5}$$
$$= -5\times 10^{5} Nm$$.


How much energy is transferred when  $$1 gm$$  of boiling water at  $$100 ^ { \circ } C$$  cools to water at  $$0^{ { \circ  } }{ C }$$  ?


1358306_1552582_ans_1e341d6977be4433a62bdc442ca71a4a.jpg


Two spherical bodies $$A$$ (radius $$6\ cm)$$ and $$B(radius\ 18\ cm)$$ are at temperature $$T_{1}$$ and $$T_{2}$$ respectively. The maximum intensity in the emission spectrum of $$A$$ is at $$500\ nm$$ and in that of $$B$$ is at $$1500\ nm$$. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by $$A$$ to that of $$B$$?

$$\lambda_{m} T = constant, \lambda_{A}T_{A} = \lambda_{B}T_{B}$$
Now the rate of total energy radiated $$E=\sigma AT^{4}$$.
$$\dfrac{E_A}{E_B}=\left(\dfrac{R_A}{R_B}\right)^{2}\times \left(\dfrac{\lambda_A}{\lambda_2}\right)^{4}$$
$$=\left(\dfrac{6}{12}\right)^{2}\times \left(\dfrac{1500}{50}\right)^{4}$$
$$=9$$


Look at Fig. 4.Mark where the heat is being transferred by conduction, by convection and by radiation. 
1847496_0bee94fba51640f3a10cc9f67f4744e8.png

1. Transfer of heat from hot plate to pan is by conduction 
2. Transfer of heat within water is by convection. 
3. Transfer of heat from hot bodies like pan, plate, burner etc. to the surroundings is radiation. 


Find the temperature $$T$$ as a function of the entropy $$S$$ of a substance for a polytropic process in which the heat capacity of the substance equals $$C$$. The entropy of the substance is known to be equal to $$S_0$$ at the temperature $$T_0$$. Draw the approximate plots $$T (S)$$ for $$C > 0$$ and $$C < 0$$. 

We know
$$S - S_0  = \displaystyle \int\limits_{T_0}^T \frac{CdT}{T} = C \ In \ \frac{T}{T_0}$$
assuming C to be a known constant
Then $$T = T_0 exp \Big( \frac{S - S_0}{C} \Big)$$
1798999_1852316_ans_acaa65b4ddb540f9bfdfc8053853008a.PNG


Calculate the temperature and density of carbon dioxide in critical state, assuming the gas to be a Van der Waals one.

$$T_{cr} = \dfrac{8}{27} \frac{a}{bR} = \frac{8}{27} \times \frac{3.62}{.043 \times .082} = 304 \ K$$

$$\rho_{cr} = \dfrac{M}{3b} = \dfrac{44}{3 \times 43} \ gm/c.c. = 0.34 \ gm/c.c.$$


Heat is always transfer ___ from bodies to ___ bodies. 

Heat is energy that flows from one object to another because of a temperature difference. The direction of heat flow is always from the $$\underline {higher temperature}$$object to the $$\underline{lower temperature}$$ object.


A nurse measures the temperature of a patient to be $$41.5^0C$$. (a) What is this temperature on the Fahrenheit scale? (b) Do you think the patient is seriously ill? Explain. 

a) By Equation 
$$T_F=\dfrac{9}{5}T_C+32=\dfrac{9}{5}(41.5^0C)+32=(74.7+32)^0F=107^0F$$
b) Yes. The normal body temperature is $$98.6^0F$$, so the patient has a high fever and needs immediate attention. 


Rub the palm of your hand on a metal surface for about 30 seconds. Place the palm of your other hand on an unrubbed portion of the surface and then on the rubbed portion. The rubbed portion will feel warmer.Now repeat this process on a wood surface. Why does the temperature difference between the rubbed and unrubbed portions of the wood surface seem larger than for the metal surface?

Rubbing a surface results in friction converting kinetic energy to thermal energy. Metal, being a good thermal conductor, allows energy to transfer swiftly out of the rubbed area to the surrounding areas, resulting i n a swift fall i n temperature. Wood, being a poor conductor, permits a slower rate of transfer, so the temperature of the rubbed area does not fall as swiftly. 


An ideal gas initially at 300 K undergoes an isobaric expansion at 2.50 kPa. If the volume increases from $$1.00 m^3$$ to $$3.00 m^3$$ and 12.5 kJ is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature?

(a) We use the energy version of the nonisolated system model. 
$$\Delta E_{int}=Q +W$$ 
where $$W = -P\Delta V$$ for a constant-pressure process 
so that
$$ \Delta E_{int} =Q - P\Delta V$$
$$ =12.5 kJ- 2.50 kPa(3.00 m^3-1.00 m^3) = 7.50 kJ$$

(b) Since pressure and quantity of gas are constant, we have from the equation of state.

$$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$$ 

and 
$$T_2=\dfrac{V_2}{V_1}T_1=(\dfrac{3.00 m^3}{1.00 m^3})(300 K)=900 K$$


The temperature difference between the inside and the outside of a home on a cold winter day is $$57.0^0F$$. Express this difference on (a) the Celsius scale and (b) the Kelvin scale.

(a) Consider the freezing and boiling points of water in each scale: $$0^0C$$ and $$100^0C; 32^0F $$and $$212^0F$$. We see that there are 100 Celsius units for every 180 Fahrenheit units: 
$$\dfrac{\Delta T_c}{\Delta T_F}=\dfrac{100^0C}{180^0F}\rightarrow \Delta T_C =\dfrac{5}{9}(\Delta T_F)=\dfrac{5}{9}(57.0)^0C=31.7^0C$$ 
(b) The Kelvin unit is the same size as the Celsius unit: 
$$T =T_c +273.15 \rightarrow \Delta T =\Delta T_c$$
$$\Delta T=\Delta T_c(\dfrac{1 k}{1 ^0C})=31.7 K (\dfrac{1 K}{1^0C}) =31.7 K$$


You need to pick up a very hot cooking pot in your kitchen. You have a pair of cotton oven mitts. To pickup the pot most comfortably, should you soak them in cold water or keep them dry?

Keep them dry. The air pockets in the pad conduct energy by heat, but only slowly. Wet pads would absorb some energy in warming up themselves, but the pot would still be hot and the water would quickly conduct a lot of energy right into you.


Convert the following temperatures to their values on the Fahrenheit and Kelvin scales: (a) the sublimation point of dry ice, $$-78.5^0C$$; (b) human body temperature, $$37.0^0C$$.

a) By Equation,
$$T_F=\dfrac{9}{5}T_C+32=\dfrac{9}{5}(-78.5)+32=-109^0F$$
And, from equation 
$$T=T_C+273.15=(-78.5+273.15)K=195K$$
b) Again,
$$T_F=\dfrac{9}{5}T_C+32=\dfrac{9}{5}(37.0)+32=98.6^0F$$
$$T=T_C+27.15=(37.0+273.15)K=310 K$$


In this experiment, you will investigate the coding of water.
Carry out the following instructions referring to Figure.
Pour $$100\,cm^3$$ of hot water into beaker A. Place the thermometer in beaker A, as shown in Figure.
(i) State one precaution that you took to ensure that the temperature reading for the hot water is as reliable as possible.
1913179_09fcfa781f6441d58ab9922824f80551.png

Precaution: Thermometer should be placed for sufficient amount of duration to record temperature reading.


In this experiment, you will investigate the coding of water.
Carry out the following instructions referring to Figure.
Suggest two factors, other than the volume and temperature of the cold water added, that affect the decrease in temperature of the hot water.
1913201_cac6bb3b0dfd402e8ed8fb9ad318d871.png

Temperature of hot water is decreased due to following reasons:
1. Evaporation: Evaporation is the process by which water changes from a liquid to a gas or vapor, it result in temperature loss of water.
2. Conduction : Conduction heat transfer is the transfer of heat by means of molecular excitement within a material without bulk motion of the matter, Here heat passes from water to beaker.


Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale )?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $$0^oC$$ and $$100^oC$$ respectively. On the absolute scale one of the fixed points is the triple-point of water, which on the kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (kelvin) scale?
(c) The absolute temperature (Kelvin scale ) T is relate to the temperature $$t_{c}$$ on the Celsius scale by $$t_{c}=T-273.15$$. Why do we have 273.15 in this relation and not 273.16?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?

(a) Triple-point has a unique temperature; fusion point and boiling point temperatures depend on pressure.
(b) The other fixed point is the absolute zero itself.
(c) Triple-point is $$0.01^oC $$, not $$0^oC$$.
(d) $$T=273.15 \times 9/5 = 491.69$$
The temperature and pressure at which a substance can exist in equilibrium in the liquid, solid, and gaseous states.It is unique and constant. The problem with Celsius scale is that boiling and melting point change with atmospheric pressure at that place. 


You work is a material testing lab and your bess tells you to increase the temperature of a sample by $$40.0^{\circ}C$$ The only thermometer you can find at your workbench reads in $$^{\circ}F$$. If The initial temperature of the sample is $$68.2^{\circ}F$$. What is its temperature in $$^{\circ}F$$, when the desired temperature increase has been achieved ?

Since $${ T }_{ F }=\dfrac { 9 }{ 5 } { T }_{ C }+32$$

Temperature increment in Fahrenheit scale is,

$$\Delta { T }_{ F }=\dfrac { 9 }{ 5 } \Delta{ T }_{ C }=\dfrac { 9 }{ 5 } 40={ 72 }^{ o }F$$

Thus, final temperature is $$T={ 68.2 }^{ o }F+{ 72 }^{ o }F={ 140.2 }^{ o }F$$

Answer is $${ 140.2 }^{ o }F$$


The fundamental interval of a thermometer $$x$$ is divided arbitrartily into $$40$$ equal and that of another thermometry into $$80$$ equal parts. If the freezing point of $$x$$ is marked $$20^{\circ}$$ and that of $$y$$ is market $$0^{\circ}$$; what is the temperature on $$x$$ when $$y$$ indicates $$70^{\circ}$$? What is the temperature in $$^{\circ}C$$? $$(55^{\circ}; 87.5^{\circ}C)$$.

If thermometer $$x$$ is divided into 40 parts and $$y$$ into 80, then 1 degree rise in temperature on $$x$$ means 2 degree rise in $$y$$.
Also if a same temperature is marked as $$0^o$$ in $$y$$ and as 20 in x, then $$x$$ is 20 steps ahead of $$y$$.
Thus $$70^{o}$$ in y means $$35^o$$ in x
However, since x is 20 steps ahead of y, 35 becomes 55.
Answer:-  $$55^o$$


What is the temperature at which we get the same reading on both the Centigrade and Fahrenheit scales?

The formulas for converting between degree Celsius 

and degree Fahrenheit are:

 °F = (°C $$\times  \dfrac{9}{5}$$) + 32

 °C = (°F - 32) $$\times  \dfrac{5}{9}$$

 To find the temperature when both are equal, 

we use an old algebra trick and just set ºF = ºC and 

solve one of the equations.

 °C = (°C$$ \times  \dfrac{9}{5}$$) + 32

 °C - (°C$$ \times \dfrac{9}{5}$$) = 32

 $$-\dfrac{4}{5} \times $$ °C = 32

 $$°C = -32 \times  \dfrac{5}{4}$$

 °C = -40

 °F = (°F $$\times  \dfrac{9}{5}$$) + 32

 °F - (°F $$\times  \dfrac{9}{5}$$) = 32

 $$-\dfrac{4}{5} \times$$  °F = 32

 °F = -32 $$\times  \dfrac{5}{4}$$

 °F = -40

 So the temperature when both the Celsius and 

Fahrenheit scales are the same is -40 degrees.



An aluminium plate fixed in a horizontal position has a hole of diameter $$2.00cm$$. A steel sphere of diameter $$2.005cm$$ rests on this hole. All the lengths refer to a temperature of $$10^oC$$. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is $$23\times 10^{-6}\ ^oC^{-1}$$ and that of steel is $$11 \times 10^{-6} \, ^oC^{-1}$$.

Given
$$d_{st} = 2.005cm$$,                    $$d_{Al} = 2.000cm$$
$$\alpha_s = 11\times 10^{-6} / ^oC$$            $$\alpha_{Al} = 23\times 10^{-6} / ^oC$$
$$d's =2.005(1 + \alpha_s \Delta T)$$               (where $$\Delta T$$ is change in temp.) 

$$\Rightarrow d's = 2.005 + 2.005 \times 11 \times 10^{-6}\Delta T$$

$$d'_{Al} = 2(1 + \alpha_{Al} \Delta T) $$
$$= 2 + 2 \times 23 \times 10^{-6}\Delta T$$

The two will slip i.e the steel ball with fall when both the diameters become equal.
So,
$$\Rightarrow 2.005 + 2.005 \times 11 \times 10^{-6} \Delta T = 2+2\times 23\times 23 \times 10^{-6}\Delta T$$

$$\Rightarrow (46 - 22.055)10^{-6} \times \Delta T = 0.005$$

$$\Rightarrow \Delta T = \dfrac{0.005\times 10^6}{23.945} = 208.81$$

Now $$\Delta T = T_2 - T_1 = T_2 - 10^oC$$  $$[\because T_1 = 10^oC \ given]$$

$$\Rightarrow T_2 = \Delta T+ T_1 = 208.81 + 10 = 281.81$$

1543615_1719832_ans_0a3faa53bd82465495bd5c1274f07747.JPG


Name the three modes of transfer of heat.

Following are the three modes of transfer of heat:

(i) Conduction: Conduction is the process of transferring heat from hot end to cold end from particle to particle of a medium.

(ii) Convection: Convection is the process of heat transfer by actual movement of medium particles.

(iii) Radiation: Radiation is the process of heat transfer in which heat directly transfers from a hot body to another cold body without affecting the medium. In radiation heat transfer, no medium is required




Explain the three modes of heat transfer .

The tightly lids or sealed bottles are first put in hot water so that the lid expands. In case of liquids , when the thermometer is put in a slightly warm water, the mercury in it rises.  When the thermometer is taken out, mercury falls again.
In cases of gases , a balloon partially inflated in a room ( cool) expands when put in warm water. Also  , a  fully inflated balloon when placed in cold water stars shrinking .
Most of the substance expands on heating while contract on cooling. Change in the temperature of a body causes change in dimensions of the body. The increase in dimensions of a body because of increase in  its temperature is known as thermal expansion . The change in length is called liner expansion. The increase in area is called area expansion . The expansion in volume is called volume expansion
When the substance is in the from of  long rod, then for small change in temperature $$(\Delta T)$$, the fractional  change in length $$(\Delta l/l_{0})$$ is directly proportional to $$\Delta T$$
Where $$\alpha $$ is called the coefficient of liner expansion and it is a characteristics of the material . From equation ( 12.3),
$$l_{0}\Delta T\alpha =\Delta l=l-l_{0}$$ 
$$\Rightarrow  l =l_{0}+l_{0}\alpha \Delta T$$
or  $$ l = l_{0}(1+\alpha \Delta T)$$
(a) Coefficient of liner expansion 
(b) Coefficient of volume expansion 
Now, considering the  fractional change in volume $$(\Delta V/V)$$, the coefficient of volume expansion $$\beta $$  is 
Where V is the initial volume of the substance . The unit of $$\alpha $$ is $$K^{-1}$$.
$$\beta $$ is also a characteristic of the substance . Generally , it depends on the temperature . It becomes constant only at a high temperature . The  value of $$\beta $$  is more for those metals whose melting point is low. The value of $$\beta $$ least solid,  more than solids in liquids , maximum in gases.
From ideal gas equation, we know that 
$$PV=nRT$$ 
At constant pressure, 
$$ P\Delta V = n R\Delta T $$ 
or, $$\Delta V = $$ \dfrac{nR}{p}\Delta T$$
or, $$\Delta V =  \dfrac{V}{T}\Delta T$$ [\because PV=nRT,\dfrac{nR}{P}=\dfrac{V}{T}]$$
$$\because \dfrac{\Delta V/V}{\Delta T}=\dfrac{1}{T}$$
or,  $$ \beta =\dfrac{1}{T}$$
Now , we will establish relation between $$\alpha $$ and $$\beta $$  
Consider a rectangular cuboid of length $$l_{1},l_{2}and l_{3}$$ and on increasing the temperature by $$\Delta T$$ , the length changes to $$l_{1},l_{2}and l_{3}$$ respectively . Then according to equation (12.4) 
$$l'_{1}=l_{1}(1+\alpha \Delta T)....(12.8)$$
$$ l'_{2}=l_{2}(1+\alpha \Delta T).....(12.9)$$
and  $$ l'_{3}=1_{3}(1+\alpha \Delta T)....(12.10)$$
Then with increase in temperature $$\Delta T$$ we represent the volume as: 
$$V+\Delta V=1'_{1}l'_{2}l'_{3}$$ $$ V+\Delta V=l_{1}(1+\alpha \Delta T)$$ $$\times l_{2}(1+\alpha \Delta T)\times l_{3}(1+\alpha \Delta T)$$ $$V+\Delta V=l_{1}l_{2}l_{3}(1+\alpha \Delta T)^{3}....(12.11)$$ $$[\because V = l_{1}l_{2}l_{3}and(a+b)^{2}=a^{2}+b^{2}+3ab(a+b)]$$ 
Neglecting the higher powers of $$\alpha$$, i e $$ \alpha ^{2}$$ and $$\alpha ^{3}$$  we have 
$$V+\Delta V\approx V(1+3\alpha \Delta T)$$ $$ or, V +\Delta V\approx V + 3\alpha V\Delta T$$ $$ or , \dfrac{\Delta V/V}{\Delta T}\approx 3\alpha $$ 
Using equation ( 12.7) , we have .
This means that the coefficient of volume expansion is thrice the coefficient of linear expansion . 
Water shows an anomalous behaviour. When it heated between $$0^{\circ}C$$and $$4^{\circ}C$$, it contracts. The volume of given water reduces when it is cooled from  the room temperature till it  reaches $$4^{\circ}C$$ . The density decreases as the volume increases below $$4^{\circ}C$$.
So, water maximum density at $$4^{\circ}C$$ . This property of water has an important effect on the enviroment. This way, animal and plant life is saved in cold region.
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Class 11 Medical Physics Extra Questions