Q. $10\ g$ of ice at $0℃$ is mixed with $100\ g$ of water at $50℃.$ What is the resultant temperature of mixture
A) $31.2℃$
B) $32.8℃$
C) $36.7℃$
D) $38.2℃$
Explanation
(d) $\theta_{mix} = \frac{m_{W}\theta_{W} - \frac{m_{i}L_{i}}{c_{W}}}{m_{i} + m_{W}} = \frac{100 \times 50 - 10 \times \frac{80}{1}}{10 + 100} = 38.2℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.