Ncert Exercises & Solutions

Three copper blocks of masses $M_{1}, M_{2} a n d M_{3}$ kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at $T_{1}, T_{2}, T_{3} (T_{1} > T_{2} > T_{3})$ . Assuming there is no heat loss to the surroundings, the equilibrium temperature $T$ is: ($s$ is the specific heat of copper)

1. $T = \frac{T_{1} + T_{2} + T_{3}}{3}$

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

3. $T = \frac{M_{1} T_{1} + M_{2} T_{2} + M_{3} T_{3}}{3 (M_{1} + M_{2} + M_{3})}$

4. $T = \frac{M_{1} T_{1} s + M_{2} T_{2} s + M_{3} T_{3} s}{M_{1} + M_{2} + M_{3}}$

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

Step 1: Equate the heat lost by one block to the heat gained by the other two blocks.

Let the equilibrium temperature of the system is T.

Let us assume that

T_{1}, T_{2} < T < T_{3} .

Heat lost by

M_{3}

= Heat gained by

M_{1}

+ Heat gained by

M_{2}

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

(where s is the specific heat of the copper material)

Step 2: Find the final temperature.

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

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

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