Question

A hot body at temperature ${\theta }_1$ is mixed with a cold body at temperature ${\theta }_2$ both having the same heat capacity, such that the rise in temperature of the cold body is equal to the fall in temperature of the hot body. The resultant temperature $\theta$ is equal to

• A. $\frac{{\theta }_1+{\theta }_2}{2}$
• B. ${\theta }_1+{\theta }_2$
• C. $\frac{{\theta }_1-{\theta }_2}{2}$
• D. ${\theta }_1-{\theta }_2$

Answer 1

The correct option is A $\frac{{\theta }_1+{\theta }_2}{2}$

$\begin{array}{cccc}Body1& {\theta }_1& S& m\\ Body2& {\theta }_2& S& m\end{array}$

Heat lost by body1$=$ Heat lost by body2

$\Rightarrow mS({\theta }_1-\theta )=mS(\theta -{\theta }_2)\Rightarrow {\theta }_1-\theta =\theta -{\theta }_2\Rightarrow {\theta }_1+{\theta }_2=2\theta \Rightarrow \theta =\frac{{\theta }_1+{\theta }_2}{2}$