Question

$45g$ of water at ${50}^{o}C$ in a breaker is cooled when $50g$ of copper at ${18}^{o}C$ is added to it. The contents are stirred till a final constant temperature is reached. Calculate this final temperature. The specific heat capacity of copper is $0.39J{g}^{-1}{K}^{-1}$ and that of water is $4.2J{g}^{-1}{K}^{-1}$. State the assumption used

• A. ${47}^{o}C$
• B. ${30}^{o}C$
• C. ${57}^{o}C$
• D. None of the above
  

Answer 1

The correct option is A ${47}^{o}C$

Heat loss by water $=$ Heat gain by copper

So, $45\times 4.2\times (50-t)=50\times 0.39\times (t-18)$

$\Rightarrow t={47}^{o}C$