Question

The specific heat capacity of ice is 2010J, of water to be 4200J and the specific latent heat of fusion of water is 3.32 x 10⁵ J. A student wants to cool m kg of water from 30^oc by adding 0.2 kg of ice initially at -3^oc. What must be the mass of the water so that the final temperature of the mixture will be 6^oc with all the ice melted?

Answer 1

Heat required to melt the ice:

$$Q_1=M\lambda.$$

Heat required to increase the temperature of ice from -3 to 0°C:

$$Q_2=Mc_i(t_0-t_{-3}).$$

As ice turned into water, find the heat required to increase the temperature of this water from 0 to 6°С:

$$Q_3=Mc_w(t_6-t_0).$$

Meanwhile m kg of water will decrease the temperature from 30 to 6С°С:

$$Q_4=mc_w(t_6-t_{30}).$$

Thermal equilibrium occurs when

$$Q_1+Q_2+Q_3+Q_4=0,\\ M(\lambda+c_i(t_0-t_{-3})+c_w(t_6-t_0))+mc_w(t_6-t_{30})=0,\\\space\\ m=M\frac{\lambda+c_i(t_0-t_{-3})+c_w(t_6-t_0)}{c_w(t_{30}-t_6)}=0.72\text{ kg}.$$