Question

How much heat is required in kJ to convert 30.7 g of water at 97.5°C to steam at 105.5°C? The boiling point of water is 100.0°C, Cm for liquid water = 75.4 J/(mol•°C), ΔHvap = 40.67 kJ/mol, and Cm for steam = 33.6 J/(mol•°C).

Answer 1

Answer on Question #72857, Chemistry / General Chemistry

How much heat is required in kJ to convert 30.7 g of water at 97.5°C to steam at 105.5°C? The boiling point of water is 100.0°C, Cm for liquid water = 75.4 J/(mol•°C), ΔHvap = 40.67 kJ/mol, and Cm for steam = 33.6 J/(mol•°C).

Solution

$Q_{\text{total}} = Q_{\text{liq}} + Q_{\text{vap}} + Q_{\text{st}}$, where $Q_{\text{liq}}$ – energy required to heat water from 97.5°C to 100°C ($\Delta T_1 = 2.5^\circ \text{C}$);

$Q_{\text{vap}}$ – energy required to evaporate water;

$Q_{\text{st}}$ – energy required to heat steam from 100°C to 105.5°C ($\Delta T_2 = 5.5^\circ \text{C}$).

$$Q_{\text{total}} = Cm_{\text{liq}}v\Delta T_1 \times \Delta H_{\text{vap}}v \times Cm_{\text{st}}v\Delta T_2$$ $$v_{\text{water}} = \frac{30.7}{18} = 1,71 \text{ (mol)}$$ $$Q_{\text{total}} = 0.0754 \times 1.71 \times 2.5 + 40.67 \times 1.71 + 0.0336 \times 1.71 \times 5.5 = 70.18 \text{ kJ}$$

Answer

70.18 kJ of heat are required to convert 30.7 g of water at 97.5°C to steam at 105.5°C.

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