Question

Figure shows two reservoirs containing two liquids of masses 4m and 2m and their specific heat capacities are S and 2S respectively. Their initial temperatures are $4T_0$ and $T_0$ respectively. The containers are joined by a conducting rod of thermal conductivity K, length $\ell$ and cross-section area A specific heat capacity of the rod is negligible.

(A) The temperature of mid-point 'P' of the rod is always constant. (B) Heat extracted from container 1 = Heat supplied to container 2. (C) After time $t_0 = \frac{2ms\ell}{KA} \ell n2$ the temperature difference between containers becomes half of the initial temperature difference. (D) After time $t_0 = \frac{ms\ell}{KA} \ell n2$ the temperature difference between containers becomes half of the initial temperature difference.

• A. (A) The temperature of mid-point 'P' of the rod is always constant.
• B. (B) Heat extracted from container 1 = Heat supplied to container 2.
• C. (C) After time $t_0 = \frac{2ms\ell}{KA} \ell n2$ the temperature difference between containers becomes half of the initial temperature difference.
• D. (D) After time $t_0 = \frac{ms\ell}{KA} \ell n2$ the temperature difference between containers becomes half of the initial temperature difference.

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

(A) The system consists of two reservoirs connected by a conducting rod. Container 1 has mass $4m$ and specific heat $S$, initially at $4T_0$. Container 2 has mass $2m$ and specific heat $2S$, initially at $T_0$. The rod has negligible heat capacity. The temperature at the midpoint $P$ of the rod is $T_P(t) = \frac{T_1(t) + T_2(t)}{2}$, where $T_1(t)$ and $T_2(t)$ are the temperatures of the containers. Due to conservation of energy, $m_1 S_1 T_1(t) + m_2 S_2 T_2(t) = \text{constant}$. Given $m_1 S_1 = (4m)(S) = 4mS$ and $m_2 S_2 = (2m)(2S) = 4mS$, we have $m_1 S_1 = m_2 S_2$. This implies $T_1(t) + T_2(t) = \text{constant}$. Therefore, $T_P(t)$ is constant.

(B) Since the rod has negligible heat capacity and there are no heat losses to the surroundings, the total heat lost by container 1 must be equal to the total heat gained by container 2, according to the principle of conservation of energy.

(C) The rate of change of temperature difference $\Delta T = T_1 - T_2$ is given by $\frac{d\Delta T}{dt} = -\frac{KA}{\ell} \left(\frac{1}{m_1 S_1} + \frac{1}{m_2 S_2}\right) \Delta T$. Substituting the values, $\frac{1}{m_1 S_1} + \frac{1}{m_2 S_2} = \frac{1}{4mS} + \frac{1}{4mS} = \frac{1}{2mS}$. The rate constant is $\alpha = \frac{KA}{\ell} \left(\frac{1}{2mS}\right) = \frac{KA}{2mS\ell}$. The time $t_0$ for the temperature difference to halve is given by $t_0 = \frac{\ln 2}{\alpha} = \frac{\ln 2}{\frac{KA}{2mS\ell}} = \frac{2mS\ell}{KA} \ln 2$.