Question

A steam engine intakes \(50\ g\) of steam at \(100°C\) per minute and cools it down to \(20°C\). If latent heat of vaporization of steam is \(540\ cal\ g^{–1}\), then the heat rejected by the steam engine per minute is ______ \(× 10^3\) \(cal.\)
(Given : specific heat capacity of water : \(1\ cal \ g^{–1} °C^{–1}\))

Answer 1

Correct Answer: 31

The problem involves calculating the heat rejected by the steam engine as steam cools from steam at \(100°C\) to water at \(20°C\). Follow these steps:
1. **Condensation**: Convert steam to water at \(100°C\). The heat released during this phase change is calculated using the formula \(Q_1 = m \cdot L\), where \(m = 50 \, \text{g}\) and \(L = 540 \, \text{cal g}^{-1}\).
\(\quad Q_1 = 50 \cdot 540 = 27000 \, \text{cal}\).
2. **Cooling**: Next, cool the water from \(100°C\) to \(20°C\). The formula \(Q_2 = m \cdot c \cdot \Delta T\) is used, where \(c = 1\, \text{cal g}^{-1} °C^{-1}\) and \(\Delta T = 100 - 20 = 80°C\).
\(\quad Q_2 = 50 \cdot 1 \cdot 80 = 4000 \, \text{cal}\).
3. **Total Heat Rejected**: Sum the heats from both steps: \(Q_{\text{total}} = Q_1 + Q_2\).
\(\quad Q_{\text{total}} = 27000 + 4000 = 31000 \, \text{cal}\).
Dividing by \(1000\) to convert to \(× 10^3\) format, we get \(31 \, × 10^3 \, \text{cal}\).
\(31\) fits within the provided range \(31,31\), confirming the solution is correct.