Question

What is the mass of water (in g) at 100 °C that can be completely boiled into liquid water at 100°C by a 88 g aluminum block at temperature 372 °C? Assume the aluminum is capable of boiling the water until its temperature drops below 100 °C.

The heat capacity of aluminum is 0.903 J g-1 °C-1 and the heat of vaporization of water at 100°C is 40.7 kJ mol-1.

Answer 1

Heat lost by aluminium =

$$88\times0.903\times(372-100) =21614.208\ J$$

So heat gain by water =

$$21614.208=m\times\frac{40.7\times10^3}{18}$$

So,

Mass$=9.56\ gm$