Question

In three experiments, a material A at a particular low temperature $T_c$ and a material B at a particular high temperature $T_H$ are placed in an isolated and insulated container. When they reach thermal equilibrium with each other (No phase change occurs), their final temperature $T_f$ is measured. The masses $m_A$ & $m_B$ and specific heats $C_A$ & $C_B$ of the material are given in the table. Assume that heat transferred is Q in the experiment. Then which of the following is/are correct :-

• A. $(T_f)_1 > (T_f)_2 > (T_f)_3$
• B. $Q_3 > Q_1, Q_2 > Q_1$
• C. $Q_2 > Q_1 > Q_3$
• D. $(T_f)_2 > (T_f)_1 > (T_f)_3$

Answer 1

Answer: (B), (D)

(B) and (D)

Answer 2

The principle of calorimetry states that the heat lost by the hotter substance equals the heat gained by the colder substance. The final equilibrium temperature $T_f$ can be calculated using the formula: $T_f = \frac{m_A C_A T_c + m_B C_B T_H}{m_A C_A + m_B C_B}$

Experiment 1: $m_A C_A = mc$, $m_B C_B = mc$. $T_{f1} = \frac{mc \cdot T_c + mc \cdot T_H}{mc + mc} = \frac{T_c + T_H}{2}$ $Q_1 = mc \left(\frac{T_c + T_H}{2} - T_c\right) = mc \left(\frac{T_H - T_c}{2}\right)$

Experiment 2: $m_A C_A = mc$, $m_B C_B = 2mc$. $T_{f2} = \frac{mc \cdot T_c + 2mc \cdot T_H}{mc + 2mc} = \frac{T_c + 2T_H}{3}$ $Q_2 = mc \left(\frac{T_c + 2T_H}{3} - T_c\right) = mc \left(\frac{2T_H - 2T_c}{3}\right) = \frac{2}{3} mc (T_H - T_c)$

Experiment 3: $m_A C_A = 2mc$, $m_B C_B = mc$. $T_{f3} = \frac{2mc \cdot T_c + mc \cdot T_H}{2mc + mc} = \frac{2T_c + T_H}{3}$ $Q_3 = 2mc \left(\frac{2T_c + T_H}{3} - T_c\right) = 2mc \left(\frac{T_H - T_c}{3}\right) = \frac{2}{3} mc (T_H - T_c)$

Comparing final temperatures (assuming $T_H > T_c$): $T_{f2} - T_{f1} = \frac{T_H - T_c}{6} > 0 \implies T_{f2} > T_{f1}$ $T_{f1} - T_{f3} = \frac{T_H - T_c}{6} > 0 \implies T_{f1} > T_{f3}$ $T_{f2} - T_{f3} = \frac{T_H - T_c}{3} > 0 \implies T_{f2} > T_{f3}$ Thus, the order of final temperatures is $T_{f2} > T_{f1} > T_{f3}$. This confirms option (D).

Comparing heat transferred, let $X = mc(T_H - T_c)$: $Q_1 = \frac{1}{2} X$ $Q_2 = \frac{2}{3} X$ $Q_3 = \frac{2}{3} X$ Since $\frac{2}{3} > \frac{1}{2}$, we have $Q_2 = Q_3 > Q_1$. This means $Q_3 > Q_1$ and $Q_2 > Q_1$. This confirms option (B).