Question

Figure shows three containers. The rightmost container contains water (heat capacity $5cal{/}^{\circ }C$), initially at temperature ${100}^{\circ }\text{C}$. The middle container contains water maintained at ${80}^{\circ }\text{C}$ with the help of a heater. The container at the left hand side contains ice at $0^{\circ }\text{C}$. There are two heat conducting rods $A$ and $B$ both having thermal resistance equal to ${10}^{\circ }\text{C}\text{s/cal}$. Disregard any heat loss in the surrounding.


Initially, what is the power of heater required to maintain the temperature of middle container at ${80}^{\circ }\text{C}$?

• A. $12\text{cal/s}$
• B. $6\text{cal/s}$
• C. $18\text{cal/s}$
• D. $20\text{cal/s}$

Answer 1

The correct option is B $6\text{cal/s}$

The thermal heat current coming into and out of the middle container are represented in the below diagram.


Since the middle container is maintained at the same temperature, it acts like a thermal junction.

i.e $i_1+i_2=i_3$

Using Fourier's law,

$\frac{\Delta T_B}{R_B}+i_2=\frac{\Delta T_A}{R_A}$

Given $R_A=R_B={10}^{0}Cs/cal$

$\frac{100-80}{10}+i_2=\frac{80-0}{10}$

$\Rightarrow i_2=6cal/s$