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Prof
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.


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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