Question

1 kg ice at -${10}^{0}C$ is mixed with 1 kg water at ${100}^{0}C$.then find equilibrium temperature and mixture content.

Answer 1

$\begin{array}{c}\left(\begin{array}{c}1kg\\ ice\\ at-{10}^{\circ }C\end{array}\right)\underset{Q_1}{\overset{absorb}{\to }}\left(\begin{array}{c}1kg\\ ice\\ at{0}^{\circ }C\end{array}\right)\underset{Q_2}{\overset{absorb}{\to }}\left(\begin{array}{c}1kg\\ water\\ at{0}^{\circ }C\end{array}\right)\\ \left(\begin{array}{c}1kg\\ water\\ at{100}^{\circ }C\end{array}\right)\underset{Q_3}{\overset{released}{\to }}\left(\begin{array}{c}1kg\\ water\\ at{0}^{\circ }\end{array}\right)\\ Q_3=m\rho \Delta \theta =1\times {10}^3\times 1\times 100\\ =100\times {10}^{3}cal\\ Q_1=m\rho \Delta \theta =1\times {10}^3\times 0.5\times 10\\ =5\times {10}^{3}cal\\ Q_2=m\rho =1\times {10}^3\times 80\\ =80\times {10}^{3}cal\\ loftheat=10\times {10}^3-85\times {10}^3\\ =15\times {10}^{3}cal\\ Now,\\ 15\times {10}^3=2\times {10}^3\times 1\times \Delta \theta \\ Hence,\\ \theta =7.5C\end{array}$