Question

$0.1m^3$ of water at ${80}^{0}C$ is mixed with $0.3m^3$ of water at ${60}^{0}C$.The final temperature of the mixture is

• A. ${65}^{0}C$
• B. ${70}^{0}C$
• C. ${60}^{0}C$
• D. ${75}^{0}C$

Answer 1

The correct option is A ${65}^{0}C$

Let the final temperature of the mixture be $t$,
Heat lost by water at ${80}^{0}C$
$ms\Delta t$
$=0.1\times {10}^3\times s_{water}\times ({80}^0-t)$
$(\because m=V\times d=0.1\times {10}^{3}kg)$
Heat gained by water at ${60}^{0}C$
$=0.3\times {10}^3\times S_{water}\times (t-{60}^0)$
According to principle of calometry
Heat lost =Heat gained
$\therefore 0.1\times {10}^3\times S_{water}\times ({80}^0-t)=0.3\times {10}^3\times s_{water}\times (t-{60}^0)$
or $({80}^0-t)=3\times (t-{60}^0)$
or $4t={260}^0$
or $t={65}^0$