A 10 g sample of ice, at - 11 is mixed with 90 ml of water at 80. calculate the final temperature of the mixture assuming that no heat is lost to the surroundings. The heat capacities of H2O (s) and H2O (I) are 2.08 and 4.18 and the heat fusion for ice is 6.

Final temperature is 71·31 °C

Explanation:

The amount of heat change in a body for a change in temperature of ΔT is m*s*ΔT

where

m is the mass of the substance

s is the heat capacity of the substance

ΔT is the temperature difference

At constant temperature the amount of heat for fusion of ice is m*L

where

m is the mass of the substance

L is the latent heat of the substance

Let the final temperature be T

As there is no heat loss to the surrounding ∴ The heat gained by the ice = The heat lost by the water

As the density of the water is 1 g/ml

∴ The mass of 90ml water will be 90 g

Heat gained by the ice when it is attained at 0°C and converted to water is

(10*2·08*11) + 10*6 = 288·8

But the heat loss by the water when water attains 0°C will be 90*4·18*80 = 30,096

As the heat loss is more therefore there will be further rise in temperature

Now the heat gain = 288·88 + 10*4·18*T

heat loss = 90*4·18 * (80-T)

Here in the case of heat loss as we are already mentioning that the heat is lost so we are taking the magnitude of the heat change otherwise it would be 90*4·18 * (T-80)

288·88 + 41·8*T = 9*41·8 * (80-T)

41·8*9*80 - 288·8 = 10*41·8*T

∴T=71·31 °C