230 cm3 of hot tea at 96 °C are poured into a very thin paper cup with 100 g of crushed ice at 0 °C. Calculate the final temperature of the "ice tea.

Final temperature of the ice tea, x = 42.73°C

Explanation:

230 cm^3 = 230 g of liquid (taking the density of tea as the same the density of water)

In cooling from 96°C to 0°

C (before freezing) the available energy is

96 * 230 = 22080 cal

The latent heat (the heat required to melt the crushed ice per gram) of fusion for water = 79.7 cal/g

Melting 100 g of crushed ice at 0°C will absorb

100 * 79.7 = 7970 cal

The retain heat of the iced Tea = (22080 - 7970) cal of heat raising the mixture temperature as follows

(22080 - 7970) = 14110 cal

The mass of the final solution 230 + 100 = 330 g solution

Final temperature = 14110 / 330 = 42.76°C

Or where specific heat of water = 4.186J/Kg°C

and the latent heat of the ice = 334J/Kg

We have

230 * (specific heat capacity of water) * (96 - x) = 100 * (specific heat capacity of water) * (0-x) + (latent heat) * 100 =

230 * 4.186 * (96 - x) = 100 * 4.186 * (x-0) + 100*334

1381.38*x = 59026.88

Final temperature of the ice tea, x = 42.73°C