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

• Take a round bottomed flask and fill it with coloured water up to the brim.
• Close the flask with a single-holed rubber stopper.
• Insert a capillary tube into the flask such that some water from the flask gets into the capillary tube and reaches a certain level.
• Heat the flask over a burner.
• Observe the level of water in the capillary tube.
• As the flask is heated, the water expands, and the level of water in the capillary tube rises.
• This is due to the thermal expansion of water.

Now stop heating the flask and observe the level of water in the capillary tube. The level of water goes down as the temperature of the water in the flask decreases. Liquids expand on heating, and contract on cooling. Thus, liquids expand on heating and contract on cooling.

Take an empty test tube and close its mouth with a single-holed rubber stopper.
Take a capillary tube in which a small amount of coloured water is trapped.
Insert it into the test tube through the single-holed rubber stopper.
Hold the test tube with a test tube holder, and heat it over a burner.
Observe the level of coloured water in the capillary tube.
On heating the test tube, the air in the test tube expands and pushes the coloured water up.
Stop heating the test tube and cool it.
If the test tube is cooled, the air contracts and the level of the coloured water comes down to its original level.

This can also be observed without heating.

• Hold the test tube tightly in your palm.
• The heat from your palm transfers to the air in the test tube.
• The air expands and pushes the coloured water up.
• Hold the test tube between your fingers.
• The air in the test tube cools and contracts.
• The level of the coloured water comes down to its original level.
• This activity confirms that gases expand even when they are exposed to very little heat.
• Thus gases expand on heating and contract on cooling.

1. As the water is heated, the density of water will decrease. So, if the objects floats at room temperature, it will start sinking as the temperature increases. Now the density of object will be more than the water.
2. The density of water is maximum at $$4^oC$$ , so the object which was floating at room temperature will continue to float but now less part of it will be submerged in water than earlier.
   

The coefficient of linear expansion, α, for a material is defined as the fractional change in length, divided by the change in temperature.

From the Ideal gas law we know that,

$$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } A=\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } }$$

Please ignore the A^ in the formula

Since pressure is constant so we have,

V₁/T₁ = V₂/T₂

=> V₂/V₁ = T₂/T₁

Given temperature T₁ = 27°C = 273 + 27 = 300k

since volume becomes double so, V₂/V₁ = 2

Hence ,

T₂/300  = 2

=> T₂ = 600k = 327°C

Hence raise in temperature = 327-27 = 300°C

Given that,

Molar mass of nitrogen, $$W=28\,g$$

Pressure is, $$P=2.46\,atm$$

Volume, $$V=10\,d{{m}^{3}}=10\,L$$

From gas equation,

  $$PV=nRT$$

$$PV=\dfrac{w}{m}RT$$

$$T=\dfrac{PVm}{WR}$$       $$R=0.0821\,Latm/K/mol$$

$$T=\dfrac{2.46\times 10\times 28}{28\times 0.0821}$$

$$T=299.63\,K$$

According to the Newton’s law o‘f cooling, the rate of loss of heat is directly proportional to the difference of temperature. Or we can say which gives a consequence about rate of fall of temperature of a body with respect to the difference of temperature of body and surroundings.

The first option would have kept water warmer because by adding hot water to cold water, the temperature of the mixture decreases. Due to this temperature difference between the mixed water in the bucket and the surrounding decreases, thereby the decrease in the rate of loss of the heat by the water.

In second option, the hot water in the bucket will lose heat quickly. So if he first attend to the work and fill the remaining bucket with cold water which already lose much heat in 5-10 minutes then the water become more colder as comparison with first case.

1)Thermal radiation emitted by a body at any temperature consists of a wide range of frequencies. The frequency distribution is given by Planck's law of black-body radiation for an idealized emitter.

2)The dominant frequency (or color) range of the emitted radiation shifts to higher frequencies as the temperature of the emitter increases. For example, a red hot object radiates mainly in the long wavelengths (red and orange) of the visible band. If it is heated further, it also begins to emit discernible amounts of green and blue light, and the spread of frequencies in the entire visible range cause it to appear white to the human eye; it is white hot. However, even at a white-hot temperature of 2000 K, 99% of the energy of the radiation is still in the infrared. This is determined by Wien's displacement law. In the diagram the peak value for each curve moves to the left as the temperature increases.

3)The total amount of radiation of all frequencies increases steeply as the temperature rises; it grows as T4, where T is the absolute temperature of the body. An object at the temperature of a kitchen oven, about twice the room temperature on the absolute temperature scale (600 K vs. 300 K) radiates 16 times as much power per unit area. An object at the temperature of the filament in an incandescent light bulb--roughly 3000 K, or 10 times room temperatureradiates 10,000 times as much energy per unit area. The total radiative intensity of a black body rises as the fourth power of the absolute temperature, as expressed by the StefanBoltzmann law. In the plot, the area under each curve grows rapidly as the temperature increases.

4)The rate of electromagnetic radiation emitted at a given frequency is proportional to the amount of absorption that it would experience by the source. Thus, a surface that absorbs more red light thermally radiates more red light. This principle applies to all properties of the wave, including wavelength (color), direction, polarization, and even coherence, so that it is quite possible to have thermal radiation which is polarized, coherent, and directional, though polarized and coherent forms are fairly rare in nature.

 On heating a body, its temperature increases.

Explanation:

• When a body is heated, energy should flow due to the law of conservation of energy (or heat). 

• It can only spread through a medium because it cannot be lost. 

• As a result, heat is absorbed by the body, raising its temperature. 

 Temperature determines the degree of hotness or coldness of a body.

Explanation:

• Temperature is the quantity which tells us the degree of hotness or coldness of a body.

• When the body is heated, its temperature rises and when the body is cooled, its temperature decreases.

(a) Fahrenheit scale.

Given, C = 40° C

As we know,

C=59(F-32)

40=59(F-32)

F-32=89=72

F=72+32

F=104° F

Hence, 40°C is equal to  104° F on the Fahrenheit scale.

(b) Kelvin scale.

(a) fahrenheit scale

As 100°C = 180°F

i.e. 100 division on celsius scale =180 division on Fahrenheit scale

So, 1 division on celsius scale =180°100°= 1.8 division on the Fahrenheit scale. 

Hence, for 1°C rise corresponding rise in fahrenheit = 1.8°F

(b) kelvin scale

Since, 100°C = 100K

i.e. 100 division on celsius scale =100 division on kelvin scale

So, 1 division on celsius scale =100°100°= 1 division on kelvin scale.

Hence, for 1°C rise corresponding rise in kelvin scale will also be 1 K.

1. A device used for heat measurement is called a calorimeter.

2. $$Construction\ of\ a\ Calorimeter$$ :

• It consists of a metallic vessel and stirrers. They are made of copper or aluminum.

• The vessel is then kept inside a wooden jacket which contains heat-insulating materials.

• The outer wooden jacket acts as a heat shield and reduces the heat loss from the inner vessel.

• The outer jacket has an opening through which a mercury thermometer is inserted into the calorimeter .

Height of Eifel tower $$= 324\,m = 32400\,cm$$ at $$15 ^\circ C$$

Temperature coefficient of linear expansion of iron $$= 11.5 \times 10^{-6}C^{-1}$$

Change in temperature $$= 30 ^\circ C - 15 ^\circ C = 15 ^\circ C$$

Change in length $$= \Delta l$$

Using formula for linear expansion of solids, we have

$$\dfrac{\Delta l}{l}=\alpha_l\Delta T$$

$$\Rightarrow \dfrac{\Delta l}{32400}=11.5 \times 10^{-6} \times 15$$

$$\Delta l=32400 \times 11.5 \times 10^{-6} \times 15=5.6cm$$

Athermanous substances- 

Substances which allow transmission of infrared radiation through them are called athermanous substances.

- For example, wood, metal, $$CO_2$$ , water vapors, etc.

Diathermanous substances - 

Substances that don't allow transmission of infrared radiation through them are called diathermanous substances.

- For example, rock salt, pure air, glass, etc.

The final volume is given as,

$${P_1}{V_1} = {P_2}{V_2}$$

$$2 \times 600 = 5 \times {V_2}$$

$${V_2} = 240\;{\rm{litres}}$$

Thus, the final volume is $$240\;{\rm{litres}}$$ .

• Radiation is the process of heat transfer in which heat is directly passed from one body at higher temperature to another body at lower temperature without affecting the medium. 

• No medium is required for the transfer of heat by radiation process.

Aim:  To show that air expands on heating.

Equipments required: 

• An empty bottle, a  rubber  balloon, a  water  bath and some water

Labelled diagram of experiment arrangement:

Procedure:

• First of all, take  an  empty  bottle.  Actually the  empty  bottle  contains  air.  

• After that attach  a  rubber  balloon to  its  neck as shown  in  Figure above.  Initially,  the  balloon  is  deflated. 

• Then place  the  bottle  in  a  water  bath  containing  boiling  water.

• After some time has passed, we will notice that the balloon has inflated as depicted in Figure.

• The reason for this is that when the air inside the bottle heats up, it expands and fills the balloon.

• Now, take  the  bottle  out  of  the  water  bath  and  allow  it  to  cool  by  itself.  

• The balloon will deflate and eventually collapse, as we will see.

• This is due to the fact that the air inside the balloon and bottle has contracted as it has cooled. The balloon deflates as the air from the balloon passes through the bottle.

Conclusion:

• The above observation shows that air expands on heating

• We feel warm when we sit in the sun. 

• We can't get heat from the sun through conduction or convection because the majority of the space between the sun and the earth is a vacuum, and both of these modes of heat transfer require a medium. 

• As a result, one must be receiving heat from the sun through the mode of radiation.

$$d=1.5m,$$
$$A = 2 \times 10^4m^2,$$
$$K = 210\,W/m/k$$ , $$\theta_1 = 373K$$ and $$\theta_2 = 263K$$ .
1. Temperature gradient
$$(\frac{\theta_{1}-\theta_{2}}{d}=\frac{373-263}{1.5}=\frac{110}{1.5}=73.33 \mathrm{K} / \mathrm{m})$$

2. Let $$\theta$$ be the temperature at a distance $$d_1 = 0.8m$$ from the hot end then temperature gradient $$= (\frac{\theta_{1}-\theta}{d^{1}})$$
i.e., $$73.33 = (\frac{373-\theta}{0.8})$$ or $$\theta =314.336\,K$$ .

3. Rate of heat transmission
$$(\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{KA}\left(\frac{\theta_{1}-\theta_{2}}{\mathrm{d}}\right))$$
$$= 210 \times 2 \times 10^{-4} \times 73.33$$

$$= 3.08\,J/s.$$