#### Question

Two bodies of masses *m*_{1} and *m*_{2} and specific heat capacities *s*_{1} and *s*_{2} are connected by a rod of length *l*, cross-sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time *t* = 0, the temperature of the first body is T_{1} and the temperature of the second body is T_{2} (T_{2} > T_{1}). Find the temperature difference between the two bodies at time *t*.

#### Solution

Rate of transfer of heat from the rod is given by

`(DeltaQ)/(Deltat) = (KA(T_2 - T_1))/l`

Heat transfer from the rod in time ΔΔ *t* is given by

`(DeltaQ)/(Deltat) = (KA(T_2 - T_1))/l Deltat ............(1)`

Heat loss by the body at temperature* **T*_{2} is equal to the heat gain by the body at temperature T_{1}

Therefore, heat loss by the body at temperature t_{2} in time Δt is given by

`DeltaQ = m_2s_2(T_2 - T_2) ....(2)`

from equation (i) and (ii)

`m_2s_2(T_2 - T_2')= (KA(T_2 - T_1))/l Delta t`

`⇒ T_2' = T_2 - (KA(T_2 - T_1))/(l(m_2s_2)) Delta t`

This gives us the fall in the temperature of the body at temperature *T*_{2}.

Similarly, rise in temperature of water at temperature *T*_{1} is given by

`T_1' = T_1 + (KA(T_2 - T_1))/(l(m_1s_1)) Delta t`

Change in the temperature is given by

`(T_2' - T_1') = (T_2 - T_1) - [(KA (T_2 - T_1))/(lm_1s_1) Deltat + (KA(T_2 - T_1))/(lm_2s_1)Delta t]`

`⇒(T_2' - T_1') - (T_2 - T_1) = - [(KA(T_2 - T

_1))/(lm_1s_1) Deltat + [(KA(T_2 - T

_1))/(lm_2s_2) Deltat]`

`rArr (DeltaT)/(Deltat)= (KA(T_2 - T_1))/l [1/(m_1s_1) + 1/(m_2 s_2)] Deltat`

`rArr 1/(T_2 - T_1) DeltaT =- (KA)/l [(m_1s_1 + m_2s_2)/(m_1s_1m_2s_2)] `

On integrating both the sides, we get

lim Δ t → 0

`int 1/(T_2 - T_1)dT = int - (KA)/l [( m_1s_1 + m_2s_2)/(m_1s_1m_2s_2) ]dt`

⇒ `In [T_2 - T_1] = - (KA)/l [( m_1s_1 + m_2s_2)/(m_1s_1m_2s_2)]t`

⇒ `(T_2 - T_1) = e^(-lamda t)`

Here , `lamda = "KA/l [ "m_1s_1 + m_2s_2"/"m_1s_1m_2s_2"]`