Question

$500g$ of water at ${100}^{\circ }C$ is mixed with $300g$ at ${30}^{\circ }C$. Find the temperature of the mixture. Specific heat of water = 4.2 J $g^{-1}$ ${}^{\circ }{C}^{-1}$).

• A. ${73.8}^{\circ }C$
• B. ${53.8}^{\circ }C$
• C. ${40}^{\circ }C$
• D. ${60}^{\circ }C$

Answer 1

The correct option is A ${73.8}^{\circ }C$

By the energy conservation, thermal energy lost by warmer water $=$ thermal energy gained by colder water

If $T_f$ be the final temperature of mixture, $m_{w}S(T_w-T_f)=m_{c}S(T_f-T_c)$ where $S=$ specific of water

or $m_w{T}_w-m_w{T}_f=m_c{T}_f-m_c{T}_c$

or $m_w{T}_w+m_c{T}_c=T_f(m_w+m_c)$

or $T_f=\frac{m_w{T}_w+m_c{T}_c}{m_w+m_c}=\frac{\left[\right(500\left)\right(100)+(300\left)\right(30\left)\right]}{500+300}={73.75}^{o}C$