Calculating Change in Entropy for Water Turning into Steam

[physgrl]

Homework Statement

Suppose that 1 kg of water, initially at 350 K, is turned into steam at 373 K. What is the change in entropy? (Lv = 2.26 x 106 J/kg; cwater = 4186 J/(kg Co)

a. 6060 J/K
b. 6070 J/K
c. 6320 J/K
d. 6330 J/K

Homework Equations

ΔS=ΔQ/T
Q=mcΔT
Q=mL

The Attempt at a Solution

heat is needed to get the water to the boiling point 100C or 373K and then heat is needed
to vaporize the water

ΔS=(mcΔT+mL)/T
ΔS=(1kg*4186J/kg*23K+1kg*2.26x10^6)/373
ΔS=6320

and the answer key says it is supposed to be 6070
what am I doing wrong?

 

[I like Serena]

Hi physgrl!

The temperature changes while heating.
This means that your formula ΔS=ΔQ/T is not right.
You need to integrate for that part of the process: ΔS=∫dQ/T

 

[physgrl]

What is the right formula?

 

[I like Serena]

For the part where the water is heating up, you should use:

ΔS=∫dQ/T

substitute dQ=m c dT

and integrate from T1=350 K to T2=373 K


Do you know how to do that?

 

[physgrl]

No sory idk how to integrate yet

 

[I like Serena]

All right. In that case you should have the following formula as a given.
\Delta S=m c \ln({T_2 \over T_1})

Do you have that somewhere on a formula sheet or something?

 

[physgrl]

Nop...i didnt have it. But with that formula how would I include the energy from the vaporization?

 

[I like Serena]

Nop...i didnt have it. But with that formula how would I include the energy from the vaporization?

Since T is constant for vaporization at T2=373 K, the change in entropy is what you already had:
\Delta S={m L \over T_2}

 

[physgrl]

Ohh ok so the total would be the sum of the mcln(t1/t2)+mL/t2

 

[I like Serena]

Almost.
You seem to have reversed the temperatures in the formula.

 

[physgrl]

Ohh ok...thanks :)

 

[I like Serena]

So... did you get the right answer?

 

[physgrl]

I did it and got 6325 but the answer key says 6070 still

 

[I like Serena]

I just calculated it myself and I also get 6325.
So I suspect your answer key is wrong.

Let me check with the specialists and I'll get back to you.

 

[physgrl]

:) ok

 

[I like Serena]

I just got a response back from my esteemed colleague and specialist in the field, vela, who says, and I quote:

The answer key appears to be wrong.

 

[physgrl]

Hehe okay thank you :)