Total change in entropy, what am I doing wrong?

[asdf188]

120 g insulated aluminum cup at 13°C is filled with 140 g of water at 58°C. After a few minutes, equilibrium is reached.

I found the final temperature to be 51°C.

But I am having trouble finding the total change in entropy.

I was using

13°C = 286.15°K
51°C = 324.15°K
58°C = 331.15°K

ΔS = ΔS_cup + ΔS_water
= Q_cup/T_avg + Q_water/T_avg
= (MCΔT)_cup/T_avg + (MCΔT)_water/T_avg
= (120g * 0.22 * (324.15°K - 286.15°K))/((324.15°K + 286.15°K)/2)
+ (140g * 1 * (324.15°K - 331.15°K))/((324.15°K + 331.15°K)/2)
=0.296567

But I am told this is the wrong answer. What am I doing wrong here?

 

[AEM]

120 g insulated aluminum cup at 13°C is filled with 140 g of water at 58°C. After a few minutes, equilibrium is reached.

I found the final temperature to be 51°C.

This is good.

I was using

13°C = 286.15°K
51°C = 324.15°K
58°C = 331.15°K

ΔS = ΔS_cup + ΔS_water
= Q_cup/T_avg + Q_water/T_avg

This last line is not good.

Have you come across this equation:

$$\Delta S = \int C \frac{dT}{T} = C ln \frac {T_2}{T_1} ?$$

 

[AEM]

120 g insulated aluminum cup at 13°C is filled with 140 g of water at 58°C. After a few minutes, equilibrium is reached.

I found the final temperature to be 51°C.

This is good.

I was using

13°C = 286.15°K
51°C = 324.15°K
58°C = 331.15°K

ΔS = ΔS_cup + ΔS_water
= Q_cup/T_avg + Q_water/T_avg

This last line is not good.

Have you come across this equation in your reading:

$$\Delta S = \int C \frac{dT}{T} = C ln \frac {T_f}{T_i}$$

C is the specific heat of your liquid or solid. $$T_f$$ is the final temp, $$T_i$$ is the initial temp. Your use of average temperature is the mistake.

 

[asdf188]

Using your formula I have 691.796 J/K from 900ln(51/13) + 4186ln(51/58) but I am still wrong in my answer. Am I supposed to convert to Kelvins? or use kcal/kg?

 

[AEM]

The "C" in front of the ln is the specific heat, so for the aluminum can that will be .215 cal/gm-C, (which is the same as cal/gm-K) and for the water 1 cal/gm-C. I'd use Kelvin. By my figuring you should get a rather small number.

 

[asdf188]

The answer is 0.94 J/K. But I have no idea how he got there.

the C for alum from our book is .22 so I got .22ln(324K/286K) + 1ln(324K/331K) = .006071 J/K

I am still missing something aren't I?

 

[Andrew Mason]

The answer is 0.94 J/K. But I have no idea how he got there.

the C for alum from our book is .22 so I got .22ln(324K/286K) + 1ln(324K/331K) = .006071 J/K

I am still missing something aren't I?

Watch the signs. The dQ for the cup is positive because heat flows into the aluminum. But the dQ for the water is negative because heat leaves the water.

AM

 

[AEM]

Also note that C is cal/gm-K. while from its definition, entropy has units of cal/K . That should be a hint to you that there's one more operation for you to do. Multiply each term by the appropriate mass and your units will come out correctly.