Question

Compute the change in internal energy when 50g of water is turned into 8.4 x 104 cm³ of steam at 100°c under normal Atm. pressure. The latent heat of vaporization of water is 2.26 x 106 J/kg.

Answer 1

$$\Delta U=Q+W=\Delta H m-p\Delta V\\ \Delta U=-(1)(84-0.05)(101.3)+2.26\cdot10^{6}(0.05)\\\Delta U=104500\ J$$