Question

A quantity of 100g of ice at 0c and 50g steam at 100 c are added to a container that has 150 g water at 30c. Determine the final temperature in the container. Ignore the container itself in your calculations.

Answer 1

weight of ice, M = 100g at 0°C
weight of steam, m = 50g at 100°C
weight of water in container , w = 150g at 30°

when ice and steam are added , heat flows through steam to ice. because we know heat flows through higher temperatures to lower temperature.

Let us assume that, all steam is condensed and form water at T °C temperature.
then, amount of heat released , H = mLv + ms(100 - T)
= 50 × 540 + 50 × 1 × (100 - T)

when all ice is melted and form water , amount of heat gained , h1 = mlf + msT
= 100 × 80 + 100 × 1 × T

water increase their temperature after gaining heat,
so, amount of heat gained by water , h2 = ms(T - 30) = 150 × 1 × (T - 30) = 150(T - 30)

heat loss = heat gain

50 × 540 + 50(100 - T) = 100 × 80 + 100T + 150(T - 30)

or, 2700 + 5(100 - T) = 800 + 10T + 15(T - 30)

or, 1900 = 10T + 15T - 450 + 5T - 500

or, 1900 = 30T - 950

or, 1900 + 950 = 30T

or, 2850 = 30T

Hence, T = 285/3 = 95°C