Question

swimmer coming out from a pool is covered with a film of water weighing about 80 grams how much heat must be supplied to evaporate the water of lantern heat of evaporation for water is 40. 79 kilo joule per mole at hundred degree celsius

Answer 1

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:

$q=n\Delta H_v_a_p$

where q is the heat energy, n is the moles and $\Delta H_v_a_p$ 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:

$80g(\frac{1mole}{18.02g})(\frac{40.79kJ}{1mole})$

= 181 kJ

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