Question

$5\, kg$ of water at $20^{\circ} C$ is added to $10\, kg$ of water at $60^{\circ} C$. Neglecting heat capacity of vessel and other losses, the resultant temperature will be nearly

• A. $35^{\circ} C$
• B. $40^{\circ} C$
• C. $47^{\circ} C$
• D. $28^{\circ} C$

Answer 1

Answer: (C)

If $T$ be the resultant temperature, then heat gained by $5\, kg \left(\right.$ at $\left.20^{\circ} C \right)$ water
$H_{\text {gain }}=m c(T-20)(\because$ given, $m=5\, kg )$
$\Rightarrow H_{\text {gain }}=5 c(T-20) \ldots$ (i)
Heat lost by $10\, kg$ water at $60^{\circ} C$ is given as
$H_{\text {loss }}=10\, c(60-T) \ldots$ (ii)
By the principle of calorimetry
$H$ gain $=H_{\text {loss }}$
$\Rightarrow 5 c(T-20)=10 c(60-T)$
$\Rightarrow T-20=120-2 T$
$\Rightarrow 3 T=140$
$T=46.67^{\circ} C =47^{\circ} C$