Question

The amount of moisture that must evaporate from a 5.0 kg body to reduce its temperature by 2°C is \( m \) g. The heat of vaporization for water at body temperature is about 580 cal/g. The specific heat capacity for the body is 0.83 cal/g°C. The value of \( m \) is:

• A. 14.3 g
• B. 19.5 g
• C. 25.4 g
• D. 35.2 g

Hint:

To calculate the moisture required to cool a body, use the specific heat for the body and the heat of vaporization for the liquid.

Answer 1

The Correct Option is B

The heat required to decrease the temperature of the body is given by: \[ Q = mc\Delta T \] where:
- \( m = 5000 \, \text{g} \) (mass of the body),
- \( c = 0.83 \, \text{cal/g°C} \) (specific heat capacity), 
- \( \Delta T = 2°C \) (temperature change). Thus, the heat required to reduce the temperature is: \[ Q = 5000 \times 0.83 \times 2 = 8300 \, \text{cal} \] Now, the heat required to evaporate \( m \) grams of water is: \[ Q_{\text{evap}} = m \times 580 \, \text{cal/g} \] Equating the two expressions for \( Q \): \[ 8300 = m \times 580 \] Solving for \( m \): \[ m = \frac{8300}{580} \approx 19.5 \, \text{g} \] Thus, the amount of moisture that must evaporate is 19.5 g.

Answer 2

This question links the heat a body loses when it cools to the heat needed to evaporate some moisture off its surface. Each option can be checked by seeing how much heat that amount of evaporating moisture would actually carry away.

1. 14.3 g: Evaporating 14.3 g of moisture at 580 cal per gram carries away \( 14.3 \times 580 \approx 8294 \) cal. The heat the body needs to lose to drop 2°C is \( 5000\,\text{g} \times 0.83\,\text{cal/g}^{\circ}\text{C} \times 2^{\circ}\text{C} = 8300 \) cal. These two numbers match almost exactly.
2. 19.5 g: Evaporating this much moisture would carry away \( 19.5 \times 580 \approx 11310 \) cal, well more heat than the 8300 cal the body actually needs to shed for a 2°C drop, so this amount overshoots.
3. 25.4 g: This would remove about \( 25.4 \times 580 \approx 14732 \) cal, nearly double the heat the body needs to lose, so this figure is far too high.
4. 35.2 g: This would remove roughly \( 35.2 \times 580 \approx 20416 \) cal, more than double the required 8300 cal, making this option even further off.

Since evaporation is what pulls heat out of the body here, the mass of moisture that evaporates has to remove exactly as much heat as the body loses while cooling by 2°C. Only 14.3 g balances that heat budget.

So the correct answer is 14.3 g.