Question

$1kg$ of cold water at $5^{\circ }C$ is mixed with $5kg$ of hot water at ${65}^{\circ }C$. Calculate the final temperature of the water when it arrives at thermal equilibrium?

• A. ${10}^{\circ }C$
• B. ${15}^{\circ }C$
• C. ${35}^{\circ }C$
• D. ${55}^{\circ }C$
• E. ${60}^{\circ }C$

Answer 1

Answer: (D)

${55}^{\circ }C$

Given : $m_1=1kg$ $T_1=5^{o}C$ $m_2=5kg$ $T_2={65}^{o}C$

Let the equilibrium temperature and specific heat of water be $T$ and $S$ respectively.

Heat released by $5kg$ water at ${65}^{o}C$ is completely absorbed by $1kg$ water at $5^{o}C$ such that net exchange of heat between system and surrounding is zero.

$\therefore$ $m_{1}S(T-T_1)+m_{2}S(T-T_2)=0$

OR $1\times S(T-5)+5\times S(T-65)=0$

OR $6T=330$ $⟹T={55}^{o}C$