Problem
The keyword “wow bao” is treated as a real-world steaming context: bao buns are commonly steamed by producing water vapor. A steamer contains 1.50 kg of liquid water initially at 22.0 °C. The water is heated to 100.0 °C, and then 0.250 kg of that water is converted to steam at 100.0 °C. Determine the total heat required.
Given (standard general chemistry values):
- Specific heat of liquid water: \(c = 4.184\,\text{J}\,\text{g}^{-1}\,^\circ\text{C}^{-1}\)
- Molar heat of vaporization of water at 100 °C: \(\Delta H_\text{vap} = 40.7\,\text{kJ}\,\text{mol}^{-1}\)
- Molar mass of water: \(M = 18.015\,\text{g}\,\text{mol}^{-1}\)
Assumptions: heat losses to the environment are neglected; heating of the steamer hardware and the bao is neglected; steam remains at 100.0 °C (no superheating).
Solution
1) Split the process into two energy parts
- Warm liquid water from 22.0 °C to 100.0 °C: use \(q = m c \Delta T\).
- Vaporize 0.250 kg of water at 100.0 °C: use \(q = n \Delta H_\text{vap}\).
2) Heat required to warm the liquid water
Convert mass to grams and compute the temperature change:
Apply \(q = m c \Delta T\):
3) Heat required to vaporize part of the water
Convert the vaporized mass to moles:
Use \(q = n\Delta H_\text{vap}\):
4) Total heat required
| Process | Expression | Value | Units |
|---|---|---|---|
| Warm liquid water (22.0 °C → 100.0 °C) | \(q_\text{heat} = m c \Delta T\) | \(4.90\times 10^2\) | kJ |
| Vaporize 0.250 kg at 100.0 °C | \(q_\text{vap} = n \Delta H_\text{vap}\) | \(5.65\times 10^2\) | kJ |
| Total (“wow bao” steam energy) | \(q_\text{total} = q_\text{heat} + q_\text{vap}\) | \(1.05\times 10^3\) | kJ |
Visualization
Final result
For the “wow bao” steaming setup, the total required heat is \(q_\text{total} \approx 1.05\times 10^3\,\text{kJ}\), with about \(4.90\times 10^2\,\text{kJ}\) for heating the water and about \(5.65\times 10^2\,\text{kJ}\) for vaporization.