Question

540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is

• A. 0°C
• B. 40°C
• C. 80°C
• D. Less than 0°C

Answer 1

Answer: (A)

0°C

Answer 2

Heat taken by ice to melt at 0°C is

$Q_{1} = mL = 540 \times 80 = 43200cal$

Heat given by water to cool upto 0°C is

$Q_{2} = ms\Delta\theta = 540 \times 1 \times (80 - 0) = 43200cal$

Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is 0°C.

Short trick : For these type of frequently asked questions you can remember the following formula

$\theta_{\text{mix}} = \frac{m_{W}\theta_{W} - \frac{m_{i}L_{i}}{c_{W}}}{m_{i} + m_{W}}$ (See theory for more details)

If $m_{W} = m_{i}$ then $\theta_{mix} = \frac{\theta_{W} - \frac{L_{i}}{c_{W}}}{2} = \frac{80 - \frac{80}{1}}{2} = 0{^\circ}C$