Two bodies of masses m1 and m2 and specific heat capacities s1 and s2 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 T1 and the temperature of the second body is T2 (T2 > T1). 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 T2 is equal to the heat gain by the body at temperature T1
Therefore, heat loss by the body at temperature t2 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 T2.
Similarly, rise in temperature of water at temperature T1 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"]`
