Question

Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.

Answer 1

The quantity of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_p \Delta T$
where m is the mass of the substance and $C_p$ is the specific heat of the substance. In our problem, we have $m=100~g=0.1~Kg$ and $C_p=4.19~KJ/(Kg K)$, which is the speficic heat of the water.
Therefore, we can find the increase in temperature $\Delta T$:
$\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C$
The initial temperature of the water was $25^{\circ}C$, so the final temperature should be 
$T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C$
So the water should be now vapor.

To be more precise, however, in the transition liquid-vapor part of the heat added to the system is used to break the bonds between molecules, and it does not increase the temperature of the system. The amount of heat needed for the transition between liquid and vapor is given by
$Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ$
where $C_L$ is the latent heat of vaporization of the water.
However, the initial amount of heat added to the water (50 KJ) is smaller than this number, therefore there isn't enough heat to complete the transition liquid-vapor. Therefore, the water will still be in the liquid-vapor change phase at a temperature of $100^{\circ}C$ (which is the temperature at which the phase change starts)