A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C are added to replace the 30 g that are evaporated. The heat of fusion of water is 333 kJ/kg, the heat of vaporization is 2256 kJ/kg, and the specific heat is 4190 J/kg · K. How many joules of energy per minute does this humidifier require?

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190

The heat of vaporization is =  2256

Mass = 0.030 kg

Energy needed to heat water to boiling point =  

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy () = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

= 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E =   +

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

this means that the temperature of both substances will be equal

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