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Question: 300 gm of water at 25°C is added to 100 gm of ice at 0°C. The final temperature of the mixture is...

300 gm of water at 25°C is added to 100 gm of ice at 0°C. The final temperature of the mixture is

A

−53∘C- \frac{5}{3}{^\circ}C

B

−52∘C- \frac{5}{2}{^\circ}C

C

– 5°C

D

0°C

Answer

0°C

Explanation

Solution

θmix=mWθW−miLiSWmi+mW=300×25−100×801100+300=−1.25oC\theta_{\text{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^{o}C

Which is not possible. Hence θmix=0oC\theta_{mix} = 0^{o}C