MCQ Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 15

After the tea is added to the thermos, the temperature of the liquid quickly falls from

$80^{o}C$ to

$75^{o}C$ as

$i$ reaches thermal equilibrium with the thermos flask
What is the heat capacity of the thermos?

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$9.5\ J/K$
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$14\ J/K$
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$95\ J/K$
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$878\ J/K$

Two identical calorimeters, each of water equivalent 100 g and volume 200 cm

$^3$ , are filled with water and a liquid. They are placed in identical constant-temperature enclosures to cool down. The temperatures are plotted at different times (the choice of units are completely arbitrary) as shown in the figure. If the density of the liquid is 800 kgm

$^{-3}$ , then its specific heat capacity is

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$8400 J/kg^o$ C
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$2100 J/kg^o$ C
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$1312.5 J/kg^o$ C
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$1680.5 J/kg^o$ C

Explanation

From the graph for the same temperature drop, (

$\Delta T$ say), the respective time taken by the liquid and water are 1 and 2 units, respectively. Average rate of heat losses for the two containers should be the same.

$\therefore (100 g \times 4200 J/kg^oC + 160 g \times s) (\Delta T/1)$

$= (100 g + 200 g) 4200 J/kg^oC (\Delta T/2)$

$\Rightarrow s = 1312.5 J/kg^o$ C

Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. Three metal rods made of copper, aluminium and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminium between the other two. The free ends of the copper and brass rods are maintained at

$100^{0}C$ and

$0^{0}C$ , respectively. The system is allowed to reach the steady-state condition. Assume there is no loss of heat anywhere.
When a steady-state condition is reached everywhere, which of the following statements is true?

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No heat is transmitted across the copper-aluminium or aluminium-brass junctions.
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More heat is transmitted across the copper-aluminium junction than across the aluminium-brass junction.
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More heat is transmitted across the aluminium-brass junction than the copper-aluminium junction.
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Equal amount of heat is transmitted at the copper-aluminium and aluminium-brass junctions.

A solid aluminium sphere and a solid lead sphere of the same radius are heated to the same temperature and allowed to cool under identical surrounding temperatures. The specific heat capacity of aluminium = $900\space J/kg^{0}C$ and that of the lead = $130 \space J/kg^{0}C$ . The density of lead = $10^{4} \space kg/m^{3}$ and that of aluminum = $2.7 \times 10^{3} \space kg/m^{3}\space kg/m^{3}$ . Assume that the emissivity of both the
spheres is the same

When the temperature of spheres T is not too different from the surrounding temperature, the radiating object obeys

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Stefan-Boltzmann law
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Newton's law of cooling
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Dulong-Petit law
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Planck's law

Explanation

Power radiated by an object to the surrounding temperature

$T_0$

$P = e \sigma A(T^{4} - T^{4}_0)$

$= \sigma eA(T^{2} + T^{2}_0) (T_0 + T_0) (T - T_0)$

$= 4e\sigma \space A \space T^{3}_0 \Delta T$

$P \space \alpha \space \Delta T$
The net power radiated is approximately proportional to the temperature difference, in agreement with Newton's law of cooling it

$(T - T_0)$ is small.

An immersion heater, in an insulated vessel of negligible heat capacity, brings 100 g of water to the boiling point from

$16^{0}C$ in 7 min. Then
Power of heater is nearly

0%
$8.4 \times 10^{3}$
0%
$84 W$
0%
$8.4 \times 10^{3}$ cal/s
0%
$20 W$

Explanation

$7 min$ , temperature of

$100 g$ of water is raised by

$(1000 - 16^{0}C) = 84^{0}C$ . The amount of the heat provided by heater

$Q_W = C_Wm_W\Delta T = (1 cal/g^{0}C)(100 g)(84^{0}C)$

$= 8.4 \times 10^{3} cal = (8.4 \times 10^{3} \times 4.186) J$

$\simeq 3.5 \times 10^{4} J$

Power of heater =

$\frac {Q_W} {t_1}$

$= \dfrac {8.4 \times 10^{3} cal} {(7 \times 60)_s} = 20 cal/s$

$= (20 \times 4.18) J/s = 83.6 W = 84 W$

A body of area $0.8 \times 10^{-2} m{-2}$ and mass $5 \times 10^{-4} kg$ directly faces the sun on a clear day. The body has an emissivity of 0.8 and specific heat of 0.8 cal/kg K. The surroundings are at $27^{0}C$ . (solar
constant = 1.4 kW/ $m^{2}$ ).

The temperature that the body would reach if it lost all its heat by radiation is

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396 K
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$296^{0}C$
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$85^{0}C$
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$85 K$

Explanation

At maximum temperature

$\dfrac {dT} {dt} = 0$

Initially

$\dfrac {dT} {dt} = 3.6^{0} C/s$

$\left [\dfrac {dT} {dt} \right ]_av = \dfrac {3.6} {2}^{0}C/s$

Rate of loss of heat

$= ms \left [\dfrac {dT} {dt} \right ]_av = \sigma A e (T^{4} - T^{4}_0)$

$T^{4} = \dfrac {\left [5 \times 10^{-4} \times 0.8 \times 4.2 \times 10^{3} \times \dfrac{3.6} {2} \right ]} {5.67 \times 10^{-8} \times 0.8 \times 10^{-2} \times 0.8} + 300^{4}$

$T = 358 K = 85^{0}C$

The only possibility of heat flow in a thermos flask is through its cork which is

$75 cm$

$^2$ in the area and

$5 cm$ thick. Its thermal conductivity is

$0.0075$ cal/cm-s-

$^o$ C. The outside temperature is

$40^o$ C and the latent heat of ice is

$80 cal/g$ . Time is taken by

$500 g$ of ice at

$0^o$ C in the flask to melt into the water at

$0^o$ C is

0%
$2.47 h$
0%
$4.27 h$
0%
$7.42 h$
0%
$4.72 h$

Explanation

Given,

$(m=500g,L=80 cal/g,k=0.0075cal/cm-s^0C,A=75cm^{2},\Delta \theta=40^{0}C,\Delta x=5cm,s=?)$

$mL = \dfrac{KA \Delta \theta t}{\Delta x}$

$\Rightarrow 500 \times 80 = \dfrac{0.0075 \times 75 \times (40 -0)t}{5}$

$\Rightarrow t = 8.9 \times 10^3$

$s = 2.47 h$

A solid aluminium sphere and a solid lead sphere of the same radius are heated to the same temperature and allowed to cool under identical surrounding temperatures. The specific heat capacity of aluminium = $900\space J/kg^{0}C$ and that of the lead = $130 \space J/kg^{0}C$ . The density of lead = $10^{4} \space kg/m^{3}$ and that of aluminum = $2.7 \times 10^{3} \space kg/m^{3} \space kg/m^{3}$ . Assume that the emissivity of both the
spheres is the same

The ration of the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the lead sphere is

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$1000:39$
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$39:1000$
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$1:1$
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$13:39$

Explanation

Let

$\dfrac{d\Delta_1}{dt}$ be the rate of fall of temperature of an aluminium sphere and

$\dfrac{d\Delta_2}{dt}$ be the rate of fall of temperature of a lead sphere.

$\dfrac {P_1} {p_2} = \dfrac {m_1 S_1 \dfrac{d\theta_1} {dt}} {m_2 S_2 \dfrac{d\theta_2} {dt}} = \dfrac {v_1d_1s_1 \dfrac{d\theta_1} {dt}} {v_2d_2s_2 \dfrac{d\theta_2} {dt}} = 1$

$\dfrac {\dfrac {d \theta_1} {dt}} {\dfrac {d \theta_2} {dt}} =\dfrac {S_2} {S_1} \dfrac{d_2} {d_1} = \dfrac{130 \times 2.7} {900 \times 10} = \dfrac {39} {1000}$

$[ \therefore V_1 = V_2]$

A room at

$20^{o}C$ is heated by a heater of resistance 20 ohm connected to 200 V mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area

$1 m^2$ and thickness 0.2 cm. Calculate the temperature outside. Thermal conductivity of glass is

$0.2 cal/ m C^{o}$ and mechanical equivalent of heat is 4.2 j/cal.

0%
$13.69^{o}C$
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$15.24^{o}C$
0%
$17.85^{o}C$
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$19.96^{o}C$

That gas cannot be liquified

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Which obeys Vander Waal's equation
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Which obeys gas equation at every temperature and pressure
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The molecules of which are having potential energy
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Which is a inert gas

According to the experiment of Ingen Hausz the relation between
the thermal conductivity of a metal rod is K and the length of the
rod whenever the wax melts

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$K/l = constant$
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$K^2/l = constant$
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$K/l^2 = constant$
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$Kl = constant$

The value of

$C_V$ for one mole of neon gas is

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$\dfrac{1}{2}R$
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$\dfrac{3}{2} R$
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$\dfrac{5}{2} R$
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$\dfrac{7}{2} R$

Which of the following statement is true

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Absolute zero degree temperature is not zero energy temperature
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Two different gases at the same temperature pressure have equal root mean square velocities
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The root mean square speed of the molecules of different ideal gases, maintained at the same temperature are the same
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Give sample of $1\,cc$ of hydrogen and $1\,cc$ of oxygen both at NTP; oxygen sample has a large number of molecules

A cubical vessel of edge

$1\,m$ and total thermal resistance (of its walls) is

$\dfrac{1}{R}$ (where, R is universal gas constant) has a small hole in one of its walls. It is kept in a a very big closed chamber whose temperature

$T_{0}$ remains constant. In the chamber and vessel, a mono-atomic gas is filled at a same constant pressure

$P_{0}.$ At time

$t = 0,$ temperature of the gas in the vessel is

$T_{1} (<2T_{0} / 3).$ When temperature of the gas in the vessel becomes

$0.8\,T_{0}.$ Rate of change of moles in it will be

0%
$\dfrac{1}{10}$
0%
$\dfrac{1}{6}$
0%
$-\dfrac{1}{10}$
0%
$-\dfrac{1}{6}$

Explanation

Gas leeks through hole

$P_{inside} = P_{outside}$

$P =$ consstant

$dQ = nc_Pdt$ ...(1)

Now heat conduction rate through wall

$= \dfrac{dq}{dt} = \dfrac{T_0 - T}{Resistance}$

$T =$ temperature of gas at that time

$t$

$\dfrac{dQ}{dt} = \dfrac{T_) - T}{1/R} = R(T_o - T)$

$dQ = R(T_0 - T)dt$ ....(2)

from (1) & (2)

$R(T_0-T)dt = nc_pdT$ ...(3)

$pv = nRT$

$o = ndT + Tdn$

$\Rightarrow ndT = -Tdn$ ....(4)

substituting (4) into (3)

$R(T_0-T)dt = -C_PTdn$

$R(T_0-T) = -C_P T \dfrac{dn}{dt}$

$\dfrac{dn}{dt}= - \dfrac{R(T_0-T)}{C_PT}$

when

$T= 0.8T_0$

$\dfrac{dx}{dt}=\dfrac{-R(T_0-0.8T_0)}{\dfrac{5}{2}R\times 0.8T_0}$

$= \dfrac{-2(T_0 - 0.8T_0)}{5\times 0.8T_0}$

$= \dfrac{-2\times 0.2}{5\times 0.8} = \dfrac{-1}{10}$

$\dfrac{dx}{dt} = \dfrac{-1}{10} = \dfrac{-1}{10}$

Rate of change of moles when

$T = 0.8 T_0$

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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