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

Solution :

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 :

$Q=mL$

$Q= 200 \times 10^{-3} \times 333.5 \times 10^3$

  $= 66700$ J

Heat required to cool the warm water is :

$Q' = MC \Delta T$

    $=500 \times 10^{-3} \times 4.18 \times 10^3 \times (20-0)$

    $=41800\ J$

From above result,

$Q > Q'$

This means that total ice could not melt.

So, the temperature must be T = 0°C

b). Q' = mL

$m=\frac{Q'}{L}$

$m=\frac{41800}{333.5 \times 10^3}$

m = 0.1253 kg

m = 125.3 g

m = 125 g

Therefore, 125 g of ice had melted.