Topic: Thermodynamics
The amount of steam (at $100^{\circ}C$) required to raise the temperature of 200 g of water from $60^{\circ}C$ to $100^{\circ}C$ is
- A. 10 g
- B. 16.8 g
- C. 20 g
- D. 14.8 g
Answer: D. 14.8 g
Explanation: The heat gained by the water must equal the heat lost by the condensing steam. Heat gained by water = $m_w \times c_p \times \Delta T = 0.2 kg \times 4.187 kJ/(kg\cdot K) \times 40 K$. The heat released by the steam during phase change is $m_s \times L$, where $L$ is the latent heat of steam ($2257 kJ/kg$). Equating these: $m_s \times 2257 = 0.2 \times 4.187 \times 40$. Solving for $m_s$ provides 14.8 g.