During exercise, the body loses heat through evaporation of sweat. If a person loses 1 kg of sweat during exercise, how much energy does the body lose through evaporation? How does the cooling effect of evaporation compare to heat loss due to specific heat capacity? (Latent heat of vaporization = 2268 × 10³ J kg⁻¹, Specific heat capacity of water = 4.2 × 10³ J kg⁻¹ K⁻¹)

Given,

Mass of sweat lost (m) = 1 kg

Latent heat of vaporization (L) = 2268 × 10³ J kg⁻¹

Specific heat capacity of water (c) = 4.2 × 10³ J kg⁻¹ K⁻¹

Now,

Energy lost by body through evaporation (Q) = mL = $1\times2268\times10^3=2268\times10^3\ J$

And,

To raise 1 kg of water by 1 °C, energy required (q) = mcΔt

On putting values,

$q=1\times4.2\times10^3\times1= 4.2\times10^3\ J$

So, evaporation of 1 kg of sweat causes heat loss equivalent to increasing the temperature of 1 kg of water by :

Equivalent temperature rise = $\dfrac{Q}{q}=\dfrac{2268\times 10^3}{4.2\times 10^3}=540\ ^oC$

The cooling effect of evaporation of 1 kg of sweat is equivalent to cooling 1 kg of water by 540°C. This shows that evaporation causes a much greater cooling effect compared to simple heating or cooling based on specific heat capacity.