Calculate change in entropy per minute.

[JoeyBob]

Homework Statement

See attached

Relevant Equations

Q=T*change in entropy

So what I did was find the change in Q per min.

Mass melted per min * latent heat capacity = Q per min = 11.5 kg /min * 3.4*10^5 J/kg = 3910000 J/min

Now the equilibrium temperature is 100 degrees Celsius or 373.15 degrees kelvin.

If I do 3910000 J/min / 373.15 K I get 10478 J/(K*min).

This seems right to me because the units are consistent. J/K are the units for entropy and the 1/min is per min. Since the question is asking for the total change in 1 min, I can multiply it by 1 min and the answer remains the same, the minute units just cancel.

But the correct answer is 3839.77, which isn't what I got.

 

[TSny]

You correctly calculated the heat absorbed by the ice-water mixture in 1 minute. But you divided this heat by the temperature of the steam-water mixture. Why?

Is there a part of the system that has a decrease in entropy? If so, what quantities do you need to know in order to calculate the decrease in entropy of this part?

Which part of the system has an increase in entropy? What do you need to know to find this increase?

 

[JoeyBob]

You correctly calculated the heat absorbed by the ice-water mixture in 1 minute. But you divided this heat by the temperature of the steam-water mixture. Why?

Is there a part of the system that has a decrease in entropy? If so, what quantities do you need to know in order to calculate the decrease in entropy of this part?

Which part of the system has an increase in entropy? What do you need to know to find this increase?

The heat absorbed by the ice-water mixture is the heat emitted by the steam-water mixture.

I divided it by temperature because that gives the right units and is consistent with the equation

Q=T*change in entropy

I suppose I need to find the temperature of the interphase maybe instead?

Here Id just assume its the average of the two temperatures, 50 celsius. But if I use 323.15 K in the eqn above I still get the wrong answer...

 

[TSny]

What is the entropy change $\Delta S_1$ of the ice-water mixture? Is $\Delta S_1$ positive or negative?

What is the entropy change $\Delta S_2$ of the steam-water mixture? Is $\Delta S_2$ positive or negative?

 

[JoeyBob]

What is the entropy change $\Delta S_1$ of the ice-water mixture? Is $\Delta S_1$ positive or negative?

What is the entropy change $\Delta S_2$ of the steam-water mixture? Is $\Delta S_2$ positive or negative?

For the ice at 273.15 K, its being heated up (dQ/dt is positive). where here dQ/dt is per min (not second)

DeltaS=dQ/dt/T=3910000/273.15 = 14314

For the steam at 373.15, its being cooled down (dQ/dt is NEGATIVE)

-3910000/373.15 = -10478

Add them up and I get a total change of 3836... close enough, rounding error or whatever.

Thanks.