Question

One kilogram of ice at 0°C is mixed with one kilogram of water at 80°C. The final temperature of the mixture is (Take : specific heat of water$= 4200Jkg^{- 1}K^{- 1}$, latent heat of ice $= 336kJkg^{- 1})$

• A. 40°C
• B. 60°C
• C. 0°C
• D. 50°C

Answer 1

Answer: (C)

0°C

Answer 2

$\theta_{\text{mix}} = \frac{m_{W}\theta_{W} - \frac{m_{i}L_{i}}{c_{W}}}{m_{i} + m_{W}}$

$\because m_{i} = m_{W}$⇒$\theta_{mix} = \frac{\theta_{W} - \frac{L_{i}}{c_{W}}}{2} = \frac{80 + 0 - \frac{336}{4.2}}{2} = 0{^\circ}C$