Q. $300\ gm$ of water at $25℃$ is added to $100\ g$ of ice at $0℃.$ The final temperature of the mixture is
A) $- \frac{5}{3}℃$
B) $- \frac{5}{2}℃$
C) $- 5℃$
D) $0℃$
Explanation
(d) $\theta_{mix} = \frac{m_{W}\theta_{W} - \frac{m_{i}L_{i}}{S_{W}}}{m_{i} + m_{W}} = \frac{300 \times 25 - \frac{100 \times 80}{1}}{100 + 300} = - 1.25℃$
Which is not possible. Hence $\theta_{mix} = 0℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.