Thermal Physics

Practice Questions:

Question 11: moderate

Consider two rods of same length and different specific heats ($S_1$, $S_2$), conductivities ($K_1$, $K_2$) and area of cross-sections ($A_1$, $A_2$) and both having temperature $T_1$ and $T_2$ at their ends. If rate of loss of heat due to conduction is equal, then
(2002)

1. $K_1 A_1 = K_2 A_2$

2. $frac{K_1 A_1}{S_1} = frac{K_2 A_2}{S_2}$

3. $K_2 A_1 = K_1 A_2$

4. $frac{K_2 A_1}{S_2} = frac{K_1 A_2}{S_1}$

View Answer

Rate of heat loss $frac{dQ}{dt} = frac{K A (T_1 - T_2)}{l}$
Given $(frac{dQ}{dt})_1 = (frac{dQ}{dt})_2 Rightarrow frac{K_1 A_1 (T_1 - T_2)}{l} = frac{K_2 A_2 (T_1 - T_2)}{l}$
$Rightarrow K_1 A_1 = K_2 A_2$

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Question 12: moderate

A cylindrical rod having temperature $T_1$ and $T_2$ at its ends. The rate of flow of heat $Q_1 text{ cal/sec}$. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat $Q_2$ will be:
(2001)

1. $4 Q_1$

2. $2 Q_1$

3. $frac{Q_1}{4}$

4. $frac{Q_1}{2}$

View Answer

$Q_1 = frac{K A Delta T}{l} = frac{K (pi r^2) Delta T}{l}$
When dimensions are doubled, $r' = 2r$, $l' = 2l$
$Q_2 = frac{K (pi (2r)^2) Delta T}{2l} = frac{4 K (pi r^2) Delta T}{2l} = 2 Q_1$

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Question 13: moderate

When $1 text{ kg}$ of ice at $0^circ text{C}$ melts to water at $0^circ text{C}$, the resulting change in its entropy, taking latent heat of ice to be $80 text{ Cal}/^circ text{C}$, is:
(2011 Pre)

1. $273 text{ cal/K}$

2. $8 times 10^4 text{ cal/K}$

3. $80 text{ cal/K}$

4. $293 text{ cal/K}$

View Answer

Entropy change $Delta S = frac{Delta Q}{T} = frac{m cdot L}{T} = frac{1000 cdot 80}{273} = 293 text{ cal/K}$

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Question 14: moderate

Thermal capacity of $40 text{ g}$ of aluminum ($s = 0.2 text{ cal/g K}$) is:
(1990)

1. $168 text{ J/K}$

2. $672 text{ J/K}$

3. $840 text{ J/K}$

4. $33.6 text{ J/K}$

View Answer

Thermal capacity $= ms = 40 cdot 0.2 = 8 text{ cal/K} = 8 cdot 4.2 text{ J/K} = 33.6 text{ J/K}$

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Question 15: moderate

$10 text{ gm}$ of ice cubes at $0^circ text{C}$ are released in a tumbler (water equivalent $55 text{ g}$) at $40^circ text{C}$. Assuming that negligible heat is taken from the surroundings the temperature of water in the tumbler becomes nearly ($L = 80 text{ cal/g}$):
(1988)

1. $31^circ text{C}$

2. $22^circ text{C}$

3. $19^circ text{C}$

4. $15^circ text{C}$

View Answer

Heat lost by tumbler = Heat gained by ice
$55 cdot (40 - T) = 10 cdot 80 + 10 cdot T$
$2200 - 55T = 800 + 10T Rightarrow 65T = 1400 Rightarrow T approx 21.5^circ text{C} approx 22^circ text{C}$

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Question 16: moderate

The unit of thermal conductivity is :
(2019)

1. $text{J m K}^{-1}$

2. $text{J m}^{-1} text{K}^{-1}$

3. $text{W m K}^{-1}$

4. $text{W m}^{-1} text{K}^{-1}$

View Answer

Thermal conductivity $K = frac{Delta Q cdot x}{A cdot Delta T cdot t}$
Unit $= frac{text{J} cdot text{m}}{text{m}^2 cdot text{K} cdot text{s}} = frac{text{W}}{text{m K}} = text{W m}^{-1} text{K}^{-1}$

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Question 17: moderate

The two ends of a metal rod are maintained at temperatures $100^circ text{C}$ and $110^circ text{C}$. The rate of heat flow in the rod is found to be $4.0 text{ J/s}$. If the ends are maintained at temperatures $200^circ text{C}$ and $210^circ text{C}$, the rate of heat flow will be:
(2015)

1. $16.8 text{ J/s}$

2. $8.0 text{ J/s}$

3. $4.0 text{ J/s}$

4. $44.0 text{ J/s}$

View Answer

Rate of heat flow $frac{dQ}{dt} = frac{K A Delta T}{L}$
Since $Delta T$ is same ($10^circ text{C}$) in both cases, the rate of heat flow will remain same i.e., $4.0 text{ J/s}$.

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Question 18: moderate

A slab of stone of area $0.36 text{ m}^2$ and thickness $0.1 text{ m}$ is exposed on the lower surface to steam at $100^circ text{C}$. A block of ice at $0^circ text{C}$ rests on the upper surface of the slab. In one hour $4.8 text{ kg}$ of ice is melted. The thermal conductivity of slab is: (Given latent heat of fusion of ice $= 3.36 times 10^5 text{ J/kg}$)
(2012 Mains)

1. $1.24 text{ J/m/s/}^circ text{C}$

2. $1.29 text{ J/m/s/}^circ text{C}$

3. $2.05 text{ J/m/s/}^circ text{C}$

4. $1.02 text{ J/m/s/}^circ text{C}$

View Answer

Heat transferred $frac{Q}{t} = frac{K A Delta T}{x}$
$frac{m L}{t} = frac{K A (100 - 0)}{x}$
$K = frac{m L x}{t A Delta T} = frac{4.8 times 3.36 times 10^5 times 0.1}{3600 times 0.36 times 100} = 1.24 text{ J/m/s/}^circ text{C}$

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Question 19: moderate

A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time $t$. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time $t$?
(2010 Pre)

1. $Q/2$

2. $Q/4$

3. $Q/16$

4. $2Q$

View Answer

$Q = frac{K A Delta T}{l} t = frac{K (pi r^2) Delta T}{l} t$
Volume is constant $Rightarrow pi r^2 l = pi (r/2)^2 l' Rightarrow l' = 4l$
$Q' = frac{K (pi (r/2)^2) Delta T}{4l} t = frac{1}{16} frac{K (pi r^2) Delta T}{l} t = frac{Q}{16}$

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Question 20: moderate

Gravitational force is required for: (2000)

1. Stirring of liquid

2. Convection

3. Conduction

4. Radiation

View Answer

Convection involves the macroscopic movement of fluid which relies on density differences. These differences in density lead to buoyant forces, which require gravity to operate.

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