How much heat needs to be imparted to 400 g of water at room temperature in order to completely evaporate it? The specific heat of vaporization of water is 2.26 * 10 to the 6th degree J / kg, the specific heat of water is 4200 J / kg * degrees Celsius. boiling point of water 100 degrees Celsius
The heat that will go to heat water from room temperature t₀ = 20 ° C to t = 100 ° C:
Q₁ = cm (t – t₀), where m is the mass, c is the specific heat capacity of water.
Heat that will go to vaporization:
Q₂ = rm, where r is the specific heat of vaporization.
Total heat required:
Q = Q₁ + Q₂ = cm (t – t₀) + rm = m (c (t – t₀) + r);
Q = 0.4 kg (4200 J / kg * gr. * (100 ° C – 20 ° C) + 2.26 * 10⁶ J / kg) =
= 0.4 (4200 * 80 + 2.26 * 10⁶) J = 1,038 400 J.
Answer: 1,038,400 J = 1.038 MJ.
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