Question

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

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

Answer 1

The correct option is A ${65}^{\circ }C$

Given volume of water $V_1=0.1m^3$ and initial temperature$T_1={80}^{0}C$ and volume $V_2=0.3m^3$ and $T_2={60}^{0}C$

Thus

$m_1{c}_p{dt}_1=m_2{c}_p{dt}_2$

as we know $m=\rho V$

$\rho \left(0.1\right)c_p(80-T_f)=\rho \left(0.3\right)c_p(T_f-60)$

$4T_f=260$

$T_f={65}^{0}C$