How Does Entropy Change Affect Heat Transfer in a Cooling Rock?

[Henry Shi]

Homework Statement

A hot rock ejected from a volcano's lava fountain cools from 1100º C to 40.0º C and its entropy decreases by 950 J/K. How much heat transfer occurs from the rock? (Source: OpenStax "College Physics for AP Students", Chapter 15.6)

Homework Equations

I used the equation ΔS_h + ΔS_c = ΔS_total, where h and c are the hot and cold states of an object, respectively.
ΔS=Q/T, where Q is joules and T is temperature Kelvin.

The Attempt at a Solution

Using the equation, we plug in the variables:
-Q_h/T_h + Q_c/T_c = -950
I set Q_h=Q_c
-Q/1373 + Q/313 = -950
Solving this equation, I got Q = 3.9 x 10⁵ Joules

However the correct answer is 8.01 x 10⁵ J
My answer is incorrect. What did I do wrong?

 

[haruspex]

ΔS=ΔQ/T only applies for small ΔQ, ΔS. As each ΔQ is lost, the temperature changes.

 

[TSny]

Hello and welcome to PF!

I used the equation ΔS_h + ΔS_c = ΔS_total, where h and c are the hot and cold states of an object, respectively.

I took a quick look at that section of the text
https://cnx.org/contents/jQSmhtXo@14.38:qmhggndY@2/Entropy-and-the-Second-Law-of-
It appears that you are using a formula that applies to the specific case of a Carnot cycle. That's not what you are dealing with in this problem.

ΔS=Q/T, where Q is joules and T is temperature Kelvin.

This formula applies to cases where the object's temperature remains constant while heat is added or removed, as in the example in the "Order to Disorder" subsection. Again, this is not what you are dealing with in this problem (as already pointed out by @haruspex). Unfortunately, the text does not appear to explain or give any example of finding the entropy change for an object that changes its temperature while heat is added or removed. This generally requires calculus. The textbook's answer is an approximate result in which you treat the object as having a constant temperature equal to the average of the initial and final temperatures. What do you get for Q if you use ΔS=Q/T where T is the average temperature?

 

[Henry Shi]

@TSny I used your method and got the following:

The average of the two temperatures is 843 K.
I then plugged into the formula ΔS=Q/T, with ΔS=950 and T=843.
950 = Q/843
Q = 800850 = 8.01 x 10⁵ J

Thank you!

 

[Chestermiller]

If the correct formula for $\Delta S$ is used (rather than the approximate formula in your reference based on the average of the two temperatures), the correct answer is 6.96 x 10⁵ J/K

 

[haruspex]

If the correct formula for $\Delta S$ is used (rather than the approximate formula in your reference based on the average of the two temperatures), the correct answer is 6.96 x 10⁵ J/K

That's using $\Delta S=\Delta Q\frac{\ln(T_f)-\ln(T_i)}{T_f-T_i}$, right? I get 6.8 x 10⁵ J/K.

For an approximate answer, I believe it would be better to use the geometric mean of the temperatures than the arithmetic mean. But why approximate?

 

[Chestermiller]

That's using $\Delta S=\Delta Q\frac{\ln(T_f)-\ln(T_i)}{T_f-T_i}$, right? I get 6.8 x 10⁵ J/K.

For an approximate answer, I believe it would be better to use the geometric mean of the temperatures than the arithmetic mean. But why approximate?

Sorry. Arithmetic error. 6.8 is right.

The correct mean to use, as evidenced by your equation, is what we engineers call the log-mean.