Question

A swimmer coming out from a pool is covered with a film of water weighing about 18 grams

Answer 1

So what is the question in it......

Answer 2

Answer:- 181 kJ

Solution:- Mass of water is 80 grams and it asks to calculate the heat required to evaporate it at 100 degree C. Latent heat of evaporation is given as 40.79 kJ per mole.

40.79 kJ per mole means 40.79 kJ of heat is required to evaporate 1 mole of water at 100 degree C. How much heat would be required to melt 80 grams of water. We need to convert the grams to moles and then multiply by latent heat of evaporation.

Also, the formula for this is written as:

where q is the heat energy, n is the moles and  is the enthalpy of vaporization or also known as latent heat of vaporization.

Molar mass of water is 18.02 g per mole. The set us could be made as:

80g(\frac{1mole}{18.02g})(\frac{40.79kJ}{1mole})80g(18.02g1mole​)(1mole40.79kJ​)

= 181 kJ

So, 181 kJ of heat is required to evaporate 80 g of water at 100 degree C.





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