Question

Some stainless steel implements, in a well insulated container, are brought into thermal equilibrium with 100 g of steam (water vapour). Initially the steam was at a temperature of 100◦C
and the implements were at a temperature of 10◦C.
[Lv(water) = 2256 kJ kg−1
c water = 4.19 kJ kg−1 K
c stainless-steel = 0.9 kJ kg−1 K
If the final equilibrium temperature of the water and steel implements is 95*C, what is the mass of the implements in kg?

Answer 1

$$Q_1 = m_{vap}L_v(H_2O)$$

$$Q_1 = 0.1 kg*2256J/kg = 225.6J$$