# How many joules of heat are needed to change 50.0 grams of ice at -15.0 °C to steam at 120.0 °C?

Dec 14, 2014

The answer is $153.7 k J$.

What you are asked to determine is the total energy required to go from ice to water, and then from water to vapor - the phase changes underwent by the water molecules.

In order to do this, you'll need to know:

Heat of fusion of water: $\Delta {H}_{f}$ = $334$ $J$/$g$;
Heat of fusion vaporization of water: $\Delta {H}_{v}$ = $2257$ $J$/$g$;
Specific heat of ice: $c$ = $2.09$ $J$/${g}^{\circ} C$;
Specific heat of water: $c$ = $4.18$ $J$/${g}^{\circ} C$;
Specific heat of steam: $c$ = $2.09$ $J$/${g}^{\circ} C$;

So, the following steps describe the overall process:

1. Determine the heat required to raise the temperature of the ice from $- {15.0}^{\circ} C$ to ${0}^{\circ} C$:

${q}_{1} = m \cdot {c}_{i c e} \cdot \Delta T = 50.0 g \cdot 2.09 \frac{J}{g {\cdot}^{\circ} C} \cdot \left({0}^{\circ} C - \left(- {15}^{\circ} C\right)\right) = 1567.5 J$

2. Determine the heat required to convert ${0}^{\circ} C$ ice to ${0}^{\circ} C$ water:

${q}_{2} = m \cdot \Delta {H}_{f} = 50.0 g \cdot 334 \frac{J}{g} = 16700 J$

3. Determine the heat required to go from water at ${0}^{\circ} C$ to water at ${100}^{\circ} C$:

${q}_{3} = m \cdot {c}_{w a t e r} \cdot \Delta T = 50.0 g \cdot 4.18 \frac{J}{g {\cdot}^{\circ} C} \cdot \left({100}^{\circ} C - {0}^{\circ} C\right) = 20900 J$

4. Determine the heat required to convert ${100}^{\circ} C$ water to ${100}^{\circ} C$ vapor:

${q}_{4} = m \cdot \Delta {H}_{v} = 50.0 g \cdot 2257 \frac{J}{g} = 112850 J$

5. Determine the heat required to go from ${100}^{\circ} C$ vapor to ${120}^{\circ} C$ vapor:

${q}_{5} = m \cdot {c}_{v a p o r} \cdot \Delta T = 50.0 g \cdot 2.09 \frac{J}{g {\cdot}^{\circ} C} \cdot \left({120}^{\circ} C - {100}^{\circ} C\right) = 2090 J$

Therefore, the total heat required is

${q}_{T O T A L} = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5} = 152696.5 J = 153.7 k J$