Question

In hot weather, a human body cools by the evaporation of sweat from its skin. The amount of water that must evaporate to cool the body by 1°C is .........% of the body mass. (Round off to two decimal places.) [Assume that latent heat of vaporization of water is \(2.25 \times 10^6 \, {J kg}^{-1}\) and specific heat capacities of both human body and liquid water are \(4.2 \times 10^3 \, {J K}^{-1} \, {kg}^{-1}\).]

Hint:

To calculate the amount of water needed for cooling via evaporation, remember to use the latent heat of vaporization and the specific heat capacity to relate the temperature change to the mass of water.

Answer 1

The heat required to cool the body by 1°C can be calculated using the specific heat capacity formula: \[ Q = m \cdot c \cdot \Delta T \] Where:
- \( Q \) is the heat removed from the body,
- \( m \) is the mass of the body,
- \( c \) is the specific heat capacity of the body,
- \( \Delta T = 1^\circ C \) is the temperature change.
Given that:
- The specific heat capacity of the human body \( c = 4.2 \times 10^3 \, {J K}^{-1} \, {kg}^{-1} \),
- The temperature change \( \Delta T = 1^\circ C \),
- The mass of the body is \( m \) (we assume 1 kg for simplicity, as we are calculating the percentage mass).
The heat required to cool the body by 1°C is: \[ Q = 1 \cdot (4.2 \times 10^3) \cdot 1 = 4.2 \times 10^3 \, {J} \] Now, to calculate the amount of water that needs to evaporate to remove this amount of heat, we use the latent heat of vaporization: \[ Q = m_{{water}} \cdot L \] Where:
- \( m_{{water}} \) is the mass of water evaporated,
- \( L = 2.25 \times 10^6 \, {J kg}^{-1} \) is the latent heat of vaporization of water.
Rearranging for \( m_{{water}} \): \[ m_{{water}} = \frac{Q}{L} = \frac{4.2 \times 10^3}{2.25 \times 10^6} \] \[ m_{{water}} = 1.87 \times 10^{-3} \, {kg} \] Now, to find the percentage of the body mass, we divide by the body mass (assuming 1 kg for simplicity): \[ {Percentage of body mass} = \frac{1.87 \times 10^{-3}}{1} \times 100 = 0.19% \] Thus, the amount of water that must evaporate is 0.18 to 0.19 percent of the body mass.