Question

How much energy should be extracted from 100 gr water vapor at 120C to transform it into water at 10C ?

Answer 1

Well first you need the specific heats of both water vapor and liquid water which are 2.08 and 4.18 J/(g*K), respectively. You also need water's heat of vaporization which is the heat needed to change liquid water to water vapor or vice versa which is 2,270 J/g.

Before you do anything, you need to convert the temperatures from deg C to Kelvin.
120 deg C = 393.15 K
100 deg C = 373.15 K
10 deg C = 283.15 K

Now use the basic heat equation:
\[Q=cM \Delta T\]
where Q is amount of heat (energy) added/lost, c is the specific heat, M is the mass of water, and Delta T is the change in temperature (T final - T initial).

To find the amount of energy lost when water vapor is brought from 393.15 K to 373.15 K:
\[Q=(2.08J/(g*K))*(100 g) *(373.15-393.15 K)=-4,160 J=-4.16 kJ\]
Now you need to factor in the heat of vaporization (you put a negative sign in because heat is lost when water goes from vapor to liquid):
\[Q=-(2,270 J/g)*(100 g)=-227,000 J=-227 kJ\]
Now you need to calculate the amount of energy lost when liquid water is brought from 373.15 K to 283.15 K:
\[Q=(4.18J/(g*K))*(100 g) *(283.15-373.15 K)=-37,620 J=-37.62 kJ\]
Finally, to get your final answer, you add these three numbers together:
\[Q=-4.16kJ-227kJ-37.62kJ=-268.78kJ\]
So 268.78 kJ of heat would need to be extracted for this process.