How much energy is required to convert 100.g of ice at 0.00 °C to water vapor at 100.00 °C?

Question

Wyatt Atkins · Accepted Answer

It takes 301 kJ of energy to change 100 g of ice at 0.00 °C into water vapor at 100.00 °C. Three heats should be taken into account: #q_1# = heat required to melt the ice to water at 0.00 °C. #q_2# = heat required to warm the water from 0.00 °C to 100.00 °C. #q_3# = heat required to vapourize the water to vapour at 100 °C. #q_1 = mΔH_(fus) = 100. g × (334 J)/(1 g)# = 33 400 J #q_2 = mcΔT = 100. g × (4.184 J)/(1 °C•g) × 100.00 °C# = 41 800 J #q_3 = mΔH_(vap) = 100. g × (2260 J)/(1 g)# = 226 000 J #q_1 + q_2 + q_3# = 33 400 J + 41 800 J + 226 000 J = 301 000 J
= 301 kJ

Cameron Andrews · Answer

undefined The energy required can be calculated using the formula:

$$ Q = m \cdot \left( \Delta H_f \, \text{for ice} + c \cdot \Delta T \, \text{for ice} + \Delta H_v \, \text{for melting} + c \cdot \Delta T \, \text{for water} + \Delta H_v \, \text{for vaporization} \right) $$

where:
- $ m $ is the mass of the substance (100.0 g),
- $ \Delta H_f $ is the heat of fusion,
- $ \Delta T $ is the change in temperature,
- $ c $ is the specific heat capacity,
- $ \Delta H_v $ is the heat of vaporization.

Given the conditions specified, this calculation requires values for $ \Delta H_f $, $ c $, and $ \Delta H_v $, which are specific to the substances involved.