Question

A beaker contains $200 g$ of water. The heat capacity of beaker is equal to that of $20 g$ of water. The initial temperature of water in the beaker is $20^{\circ} C$. If $440 g$ of hot water at $92^{\circ} C$ is poured in, the final temperature, neglecting radiation loss, will be

• A. $58^{\circ} C$
• B. $68^{\circ} C$
• C. $73^{\circ} C$
• D. $78^{\circ} C$

Answer 1

Answer: (B)

Heat lost $=(440 g )\left(1\right.$ cal $\left. g ^{-1}{ }^{\circ} C ^{-1}\right)(92-T)^{\circ} C$
Heat gained $=(200+20) g\left(1 \operatorname{calg}^{-1}{ }^{\circ} C ^{-1}\right) \cdot(T-20){ }^{\circ} C$
$\Rightarrow 440(92-T)=220(T-20)$ or $2(92-T)=T-20$
or $(2 \times 92)-2 T=T-20$
$ \Rightarrow 3 T=(184+20)$
$T=\frac{204}{3} $
$\Rightarrow T=68^{\circ} C$