Question

What is the generated entropy (in kJ/K) when 1 kg of liquid water at 25°C is heated to saturated vapor at 100°C in an oven that is at 150°C. Hint: the effect of pressure on the entropy of liquids or solids is usually negligible. Now assume you are trying to heat 1 kg of liquid water at 25°C to saturated vapor at 100°C, but you want to do it in an oven that is at 80°C. This is obviously impossible to do, but calculate what would be the generated entropy (in kJ/K) to demonstrate it is not possible.

Answer 1

Hi
For 1 kg of water.
When we heat from 25 o C to 100 o C.

Let us assume that the heat is added reversibly.
Total heat added = heat added to raise temperature to 100 oC+ latent heat of vaporization.

=mcp*(Tf-Ti)+mL
cp for the liquid water can be taken as 4.18 kJ/kg.

Latent heat of vaporization = 2260 Kj/kg for water

So total heat = 4.18*(100-25)+2260
=313.5+2260
=2573.5 kJ

Change in entropy = dq /T
= 2573.5/373 =6.89 kJ /mole

For the oven dq= -2573.5 kJ
So the entropy = -6.89 kj/mole

Net change in entropy =6.89-6.89 =0

(It is zero as the process is reversible)

We can see that if the oven is af 80 o C .
When we heat the substance temperature gradually increases to 100 o C ,so at 80 a thermal equlibrium will be reached and thus the temperature of both will start increasing together.

Mathematically
It the oven is at 80 oC .

Heat released = m*cp(80- 25)
-4.18*55
= -230 kj

The temperature of the oven is increasing only up to 80 o C ,so the entropy of oven will increase only up to that temperature


Change in entropy = -230/100 = -2.3 kJ/mole

Net entropy change = 6.89-2.3 =4.59 kj/mole
Change in the entropy can not be positive thus we can say that this process is not possible.

Hope this helps


so it is not possible to heat up to 100 o C .

Hope this helps
Feel free to comment in the case of any doubt.
Thank you