Question

A 200 g piece of ice at 0°C is placed in 500 g of water at 20°C. The system is in a container of negligible heat capacity and insulated from its surroundings.

a. What is the final equilibrium temperature of the system?
b. How much of the ice melts?

Answer 1

Given data :

Mass of ice, m = 200 g

Mass of water, M = 500 g

Temperature of ice = 0°C

Temperature of water = 20°C

We known ;

The latent heat of fusion of water is
`L=333.5 \ kJ/kg`

The specific heat of water, C = 4.18 kJ/kg K

a). Heat required to change the ice into water is :

`Q=mL`

`Q= 200 * 10^(-3) * 333.5 * 10^3`

`= 66700` J

Heat required to cool the warm water is :

`$Q' = MC \Delta T$`

`$=500 * 10^(-3) * 4.18 * 10^3 * (20-0)$`

`$=41800\ J$`

From above result,

`Q > Q'`

This means that total ice could not melt.

So, the temperature must be T = 0°C

b). Q' = mL

`$m=(Q')/(L)`

`$m=(41800)/(333.5 * 10^3)$`

m = 0.1253 kg

m = 125.3 g

m = 125 g

Therefore, 125 g of ice had melted.