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Sagot :
Answer:
2232.5 joules/gm
Explanation:
The specific heat of water and steam (water vapor) are 4.18
J/gºC and 2.10 J/gºC respectively. If 127 kJ of heat is need to
heat a 54.0 g sample of water from 74.20 °C to water vapor at
105.50 °C, what is the heat of vaporization of water in J/g?
Assume the boiling point of water is 100.00 °C.
we have to raise 54.0 gm of water from 74.20 °C to 100.00 °C, it will cost
4.18 X54 X (100-74.20) =4.18 X 54 X 25.8 = 5823.6 joules.
once we vaporize the water, we raise the temperature from 100.00 to 105.50C. this will cost 54 X 2..1 X (105.5 - 100.00) = 54 X 2.1 X 5.5 =
623.7 joules.
the total thermal cost of the whole process from water at 74.20 °C to steam
at 105.5 °C is 127,000 joules
the cost for vaporizing 54...0 gm of 100.00 °C. liquid water to steam is
127,000 joules - 623.7 joules -5823.6 joules = 120552.7 joules
54.0 gm, therefor the heat o vaporize water per gram is
120552.7/54 = 2232.5 joules/gm
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