Question

Two blocks with heat capacities $C_1$ and $C_2$ are connected by a rod of length l, cross-sectional are A and heat conductivity K. Initial temperature difference between the two blocks is $T_0$. Assuming the entire system to be isolated from surroundings, heat capacity of the rod to be negligible. The temperature difference between the blocks as a function of time is.

• A. $T_{0}exp\left(\frac{-KA(C_1+C_2)t}{C_1{C}_2}\right)$
• B. $T_{0}exp\left(\frac{-KA(C_1+C_2)}{C_1{C}_2}\right)$
• C. $T_{0}exp\left(\frac{KA(C_1+C_2)t}{C_1{C}_2}\right)$
• D. $T_{0}exp\left(\frac{-KA(C_1+C_2)t^2}{C_1{C}_2}\right)$

Answer 1

Answer: (A)

$T_{0}exp\left(\frac{-KA(C_1+C_2)t}{C_1{C}_2}\right)$

Let T be the temperature difference between two blocks at time t.

Heat transferred per second,

$\frac{dQ}{dt}=\frac{KAT}{l}$ $\dots$ (i)

Also, $dT={dT}_1+{dT}_2$ $\dots$ (ii)

Heat lost by one block is equal to the heat gained by the

other, $C_1{dT}_1=C_2{dT}_2$ $\dots$ (iii)

From equations, (ii) and (iii), we get

$dT=\left(\frac{C_1+C_2}{C_2}\right){dT}_1$ $\dots$ (iv)

If a block loses heat, $dQ=-C_1{dT}_1$

From equation (i), $\frac{dQ}{dt}=-C_1\frac{{dT}_1}{dt}=\frac{KAT}{l}$ (v)

From equations (iv) and (v), we get

$\frac{-C_1{C}_2}{C_1+C_2}\frac{dT}{dt}=\frac{KA}{l}T$

$\frac{dT}{T}=\frac{-KA(C_1+C_2)}{C_1{C}_2}dt$

or ${\int }_{T_0}^T\frac{dT}{T}=\frac{-KA(C_1+C_2)}{C_1{C}_2}{\int }_0^{t}dt$

In $\frac{T}{T_0}=\frac{-KA(C_1+C_2)}{C_1{C}_2}t$

or $T=T_{0}exp\left(\frac{-KA(C_1+C_2)t}{C_1{C}_2}\right)$