Compute the heat of vaporization at 37 0C. Find the data you need in a good source and cite it properly. Give a qualitative explanation for the difference between the values at 250Cand 37 0C.

at 100g and 37oC

Hv=154882J/Kg

at 100g and 25oC

Hv=104650J/Kg

Explanation:

The amount of energy required is called the heat of vaporization. The heat of vaporization is different for all substances, but is a constant for each individual substance.

Hv = heat of vaporization

q = heat

m = mass

for this question we wil bel taking a sample of 100g of water. But we know the heat q. so,

To calculate the amount of heat released in a chemical reaction, use the equation Q = mc ΔT, where Q is the heat energy transferred (in joules), m is the mass of the liquid being heated (in kilograms), c is the specific heat capacity of the liquid (joule per kilogram degrees Celsius), and ΔT is the change in .

Specific heat capacity of water = c = 4,186 J/kg degrees C

At 37oC

ΔT=(37-0)=37

Q=100x4186x37= 15,488,200J

Therefore ,

Hv=15488200/100=154882J/Kg

At 25oC

ΔT=(25-0)=25

Q=100x4186x25= 10,465,000J

Hv=10465000/100=104650J/Kg

So we can see that at high temperatures a great deal of heat is dispensed.

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