Two bodies having masses m_1 and m_2 along with specific heat...

Let the temperatures of the two bodies at time $t$ be $\theta_1$ and $\theta_2$ respectively.

The rate of heat flow from the second body to the first body is given by:

$\dfrac{dQ}{dt} = \dfrac{KA(\theta_2 - \theta_1)}{l}$

The heat gained by the first body and lost by the second body changes their temperatures. The rates of change of temperatures are:

$\dfrac{d\theta_1}{dt} = \dfrac{1}{m_1 s_1} \dfrac{dQ}{dt} = \dfrac{KA(\theta_2 - \theta_1)}{l m_1 s_1}$

$\dfrac{d\theta_2}{dt} = -\dfrac{1}{m_2 s_2} \dfrac{dQ}{dt} = -\dfrac{KA(\theta_2 - \theta_1)}{l m_2 s_2}$

The rate of change of the temperature difference $\Delta \theta = \theta_2 - \theta_1$ is:

\dfrac{d(\Delta \theta)}{dt} = \dfrac{d\theta_2}{dt} - \dfrac{d\theta_1}{dt} = -\dfrac{KA}{l} (\theta_2 - \theta_1) \left(\dfrac{1}{m_1 s_1} + \dfrac{1}{m_2 s_2} \right)

\dfrac{d(\Delta \theta)}{dt} = -\dfrac{KA(m_1 s_1 + m_2 s_2)}{l m_1 m_2 s_1 s_2} \Delta \theta

Let $\lambda =$\dfrac{KA(m_1 s_1 + m_2 s_2)}{l m_1 m_2 s_1 s_2}. The equation becomes:

\dfrac{d(\Delta \theta)}{dt} = -\lambda \Delta \theta

Integrating both sides from $t = 0$ to $t$, where the initial temperature difference is $T_2 - T_1$:

$\int_{T_2 - T_1}^{\Delta \theta}$\dfrac{d(\Delta \theta)}{\Delta \theta} = -\int_0^t \lambda dt$<br><br>$\ln \left( \dfrac{\Delta \theta}{T_2 - T_1} \right) = -\lambdat$<br><br>$\Delta \theta = (T_2 - T_1) e^{-\lambda t}