50g of ice at 0°C is mixed with 50g of water at 80°C, what will be the final temperature of a mixture in

degrees C?

0° C

Explanation:

Given that

Mass of ice, m = 50g

Mass of water, m (w) = 50g

Temperature of ice, T (i) = 0° C

Temperature of water, T (w) = 80° C

Also, it is known that

Specific heat of water, c = 1 cal/g/°C

Latent heat of ice, L (w) = 89 cal/g

Let us assume T to be the final temperature of mixture.

This makes the energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice (water) to raise its temperature at T° C = heat lost by water to reach at T° C

m (i). L (i) + m (i). c (w) [T - 0] = m (w). c (w) [80 - T], on substituting, we have

50 * 80 + 50 * 1 (T - 0) = 50 * 1 (80 - T)

4000 + 50T = 4000 - 50T

0 = 100 T

T = 0° C

Thus, the final temperature is 0° C