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.