Question

$5kg$ of water at $20^{\circ }C$ is added to $10kg$ 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 $5kg($ at $20^{\circ }C)$ water
$H_{\text{gain }}=mc(T-20)(\because$ given, $m=5kg)$
$\Rightarrow H_{\text{gain }}=5c(T-20)\dots$ (i)
Heat lost by $10kg$ water at $60^{\circ }C$ is given as
$H_{\text{loss }}=10c(60-T)\dots$ (ii)
By the principle of calorimetry
$H$ gain $=H_{\text{loss }}$
$\Rightarrow 5c(T-20)=10c(60-T)$
$\Rightarrow T-20=120-2T$
$\Rightarrow 3T=140$
$T=46.67^{\circ }C=47^{\circ }C$