Question

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.

Answer 1

The temperature difference between the two bodies at time t is given by

$d T=(T 2-T 1) e^{-\gamma t}$

Explanation:

The temperature difference between the two bodies at time t:

Step 1:

Rate of transfer of heat from the rod is given by the following formula

        $\frac{\Delta Q}{\Delta t}=\frac{K A(T 1-T 2)}{l}$          -------> (1)

Where T1 and T2 are the temperature of the body

From eq 1 we can write            

      $\Delta Q=\frac{K A(T 1-T 2) \Delta t}{l}$    ---------->(2)

We know that Heat loss at T2 = Heat gain T1

Heat loss of body at T2 :

      $\Delta Q=m 2 s 2\left(T 2-T^{\circ} 2\right)$ ----->(3)  

Step 2:    

Subsituting the values in equation  

$m 2 s 2\left(T 2-T^{\circ} 2\right)=\frac{K A(T 1-T 2) \Delta t}{l}$

T°2 is given by the following,

$T^{\circ} 2=T 2-\frac{K A(T 1-T 2) \Delta t}{l}$

Step 3:

Raise in temperature of water at T1  is given by,

$m 1 s 1\left(T^{\circ} 1-T 1\right)=\frac{K A(T 1-T 2) \Delta t}{l}$

$T^{\circ} 1=T 1+\frac{K A(T 1-T 2) \Delta t}{l m 1 s 1}$

$T^{\circ} 2-T^{\circ} 1=T 2-T 1-\frac{K A(T 1-T 2) \Delta t}{l}-\frac{K A(T 1-T 2) \Delta t}{l m 1 s 1}$

$\left(T^{\circ} 2-T^{\circ} 1\right)-(T 2-T 1)=-\frac{K A(T 1-T 2) \Delta t}{l}\left[\frac{1}{m 2 s 2}-\frac{1}{m 1 s 1}\right]$                ( $\Delta t=0$)

$\frac{\Delta T}{\Delta t}=\frac{-K A}{l} \frac{m 1 s 1-m 2 s 2}{m 1 m 2 s 1 s 2}(T 1-T 2)$

$\frac{d T}{T 1-T 2}=\frac{-K A}{l} \frac{m 1 s 1-m 2 s 2}{m 1 m 2 s 1 s 2} d t$

$\ln (T 1-T 2)=\frac{-K A}{l} \frac{m 1 s 1-m 2 s 2}{m 1 m 2 s 1 s 2} t+C$                                    ( $\Delta t=T 2-T 1$ )      

$\ln (T 2-T 1)=C$

$\ln d T=\frac{-K A}{l} \frac{m 1 s 1-m 2 s 2}{m 1 m 2 s 1 s 2}+\ln (T 2-T 1)$

$\ln \frac{d T}{\tau_{1-T 2}}=-\gamma t$                                            where $\gamma=\frac{K A}{l} \frac{m 1 s 1-m 2 s 2}{m 1 m 2 s 1 s 2}$

$d T=(T 2-T 1) e^{-\gamma t}$          

Thus the temperature difference between the bodies is given by  $d T=(T 2-T 1) e^{-\gamma t}$                              

Answer 2

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