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

The heat of vaporization of water Qw is equal to heat gained by aluminum Qal.

$Qw=Qal$

$Mw*cv*dTw=mal*cp*dTal$

where; Mw is the mass of water, cv is the heat of vaporization of water, dT is the change in temperature, mal is the mass of aluminium, cp is the heat capacity of aluminum.

$Mw=(mal*cp*dTal)/ (cv*dTw$

Mw = (88*0.903*372)/ (40.7*100)

Mw =7.26g