Ncert Exercises & Solutions

Three copper blocks of masses

M_{1}, M_{2}

$M_{1} , M_{2}$ and

M_{3} kg

$M_{3}~\text{kg}$ respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at

T_{1}, T_{2}, T_{3} (T_{1} > T_{2} > T_{3}) .

$T_{1} , T_{2} , T_{3} \left(\right. T_{1} > T_{2} > T_{3} \left.\right).$ Assuming there is no heat loss to the surroundings, the equilibrium temperature

T

$T$ is:
(

s

$s$ is the specific heat of copper)

1.	$T = \dfrac{T_{1} + T_{2} + T_{3}}{3}$
2.	$T = \dfrac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}$
3.	$T = \dfrac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{3 \left(\right. M_{1} + M_{2} + M_{3} \left.\right)}$
4.	$T = \dfrac{M_{1} T_{1} s + M_{2} T_{2} s + M_{3} T_{3} s}{M_{1} + M_{2} + M_{3}}$

Hint: The heat lost by one block will be equal to the heat gained by the other two blocks.

Step 1: Find the value of heat lost by one block and the heat gained by the other two blocks.

Let the equilibrium temperature of the system is

T .

$T.$

Let us assume that

𝑇_{1}, 𝑇_{2} < 𝑇 < 𝑇_{3} .

$𝑇 _1 , 𝑇 _2 < 𝑇 < 𝑇 _3 .$

Since there is no heat loss to the surroundings, the net heat exchange must be zero: Heat gained+Heat lost=0
The heat exchanged by each block is given by:

Q = m s Δ T

$Q=msΔT$
where

m

$m$ is the mass,

s

$s$ is the specific heat, and

Δ T

$\Delta T$ is the change in temperature.
For block 1 (initially at

T_{1}

$T_1$ ):
Heat lost

= M_{1} s (T_{1} - T) . . . (1)

$=M_1 s (T_1 - T)~~~...(1)$
For block 2 (initially at

T_{2}

$T_2$ ):
Heat gained/lost

= M_{2} s (T - T_{2}) . . . (2)

$=M_2 s (T - T_2)~~~...(2)$
(This block can either gain or lose heat depending on the equilibrium temperature

T .

$T.$ )
For block 3 (initially at

T_{3}

$T_3$ ):
Heat gained

= M_{3} s (T - T_{3}) . . . (3)

$= M_3 s (T - T_3)~~~...(3)$

Step 2: Find the equilibrium temperature

T .

$T.$
Apply energy conservation, since the total heat exchange is zero.

M_{1} s (T_{1} - T) = M_{2} s (T - T_{2}) + M_{3} s (T - T_{3})

$M_1 s (T_1 - T) = M_2 s (T - T_2) + M_3 s (T - T_3)$
Solve for the equilibrium temperature

T,

$T,$ divide throughout by

s

$s$ (since it is the same for all blocks) now expand the terms.

\Rightarrow T (M_{1} + M_{2} + M_{3}) = M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}

$\Rightarrow T (M_1 + M_2 + M_3) = M_1 T_1 + M_2 T_2 + M_3 T_3$

\Rightarrow T = \frac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{M_{1} + M_{2} + M_{3}}

$\Rightarrow T = \frac{M_1 T_1 + M_2 T_2 + M_3 T_3}{M_1 + M_2 + M_3}$

Hence, option (2) is the correct answer.

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