Question

A jar of tea is placed in sunlight until it

reaches an equilibrium temperature of 33.4◦C .

In an attempt to cool the liquid, which has a

mass of 184 g , 103 g of ice at 0.0◦C is added.

At the time at which the temperature of the

tea is 32.6◦C , find the mass of the remaining ice in the jar. The specific heat of water

is 4186 J/kg ·◦ C . Assume the specific heat

capacity of the tea to be that of pure liquid

water.

Answer in units of g.

Answer 1

The heat balance equation says

$$c_wm_w(t_i-t_f)=\lambda_{ice}\Delta m_{ice}.$$

Hence

$$\Delta m_{ice}=\frac{c_wm_w(t_i-t_f)}{\lambda_{ice}}\\ =\frac{4.186\times 184\times(33.4-32.6)}{334}=1.84\:\rm g.$$

The mass of the remaining ice in the jar

$$m=103-1.84=101\:\rm g.$$