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A hiker, caught in a rainstorm might absorb 1 liter of water in her clothing. if it is windy so that this amount of water is evaporated quickly at 20�c, how much heat would be required for this process?

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Water evaporates at 100⁰C

So change in temperature = 100-20 = 80⁰C

Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg

Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg

So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ

So amount of heat require to evaporate water = 334.88 kJ

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