# Calculate Entropy Change for 1.6kg Water at 34°C and 2.0kg Water at 58°C

• Jayhawk1
In summary, the question involves mixing two different amounts of water at different temperatures in a well-insulated container. The final temperature can be calculated using the formula for heat transfer. The approximate change of entropy can be found by using the average temperature of the warm and cool water in the entropy formula. If negative answers are obtained, there is an error in the calculations for the entropy change. Integrating may also be necessary for more accurate results.
Jayhawk1
Here is the question:

8) [1.0/3.0] 1.6 kg of water at 34oC is mixed with 2.0 kg of water at 58o. in a well insulated container. a) What is the final temperature of the water (in degrees Celsius)? b) What is the approximate change of entropy? (Use the average temperature for the entropy calculation.)

Part a was easy enough... I got 47.3 oC

I've tried calculating part b in multiple ways, and I am getting a very small number- which is what my answer is actually suppose to be- BUT - it is not right. Any help?? Thanks.

hmm...well you know that deltaH = qt and qt/t = delta S...give that a try?

Nothing I've trired has worked... I keep coming up with a negative number... I need some definitive direction.

Last edited:
Jayhawk1 said:
Nothing I've trired has worked... I keep coming up with a negative number... I need some definitive direction.

$$\Delta S = \frac{\Delta Q}{T}$$

S is entropy, Q is heat, T is temperature. You are asked for an approximate value for the change in entropy using an average termperature. I interpret that to mean you should use the average temperature of the warm water to calculate its entropy loss and the average temperature of the cool water to calculate its entropy gain. There will be a net gain.

Still not working... what else should I try??

Can someone be a little more specific... I am getting really frustrated. This is the last question on the last homework assignment for the year and I keep getting it wrong. Not to mention I don't have that many tries on it left. Please help me!

Jayhawk1 said:
Can someone be a little more specific... I am getting really frustrated. This is the last question on the last homework assignment for the year and I keep getting it wrong. Not to mention I don't have that many tries on it left. Please help me!

How about showing us exactly what you did. I thought the last thing I told you was quite specific. What did you get for the amount of heat transferred from the warm water to the cold water, the entropy lost by the warm water, and the entropy gained by the cold water. If you are still getting negative answers for the net entropy change, you are doing something wrong in your calculations.

Jayhawk1 said:
Can someone be a little more specific... I am getting really frustrated. This is the last question on the last homework assignment for the year and I keep getting it wrong. Not to mention I don't have that many tries on it left. Please help me!

You cannot continue complaining that you are not getting the right answer. The way you have gone about with this appears as if you want someone to do the work for you.

Show EXACTLY what you did to get the wrong answer. Only then can someone point out what you did wrong. You learn MORE from mistakes like this than from the ones you did correctly.

Zz.

Same problem for me. I've tried getting Q by computing specific heat * g * delta T, but no luck with that.

squib said:
Same problem for me. I've tried getting Q by computing specific heat * g * delta T, but no luck with that.

What answer are you getting, how are you getting it, and what (if you know) are you supposed to get?

Jayhawk1 said:
Here is the question:

8) [1.0/3.0] 1.6 kg of water at 34oC is mixed with 2.0 kg of water at 58o. in a well insulated container. a) What is the final temperature of the water (in degrees Celsius)? b) What is the approximate change of entropy? (Use the average temperature for the entropy calculation.)

Part a was easy enough... I got 47.3 oC

I've tried calculating part b in multiple ways, and I am getting a very small number- which is what my answer is actually suppose to be- BUT - it is not right. Any help?? Thanks.
Since the temperature changes as the heat is absorbed/lost, you have to integrate.

AM

Andrew Mason said:
Since the temperature changes as the heat is absorbed/lost, you have to integrate.

AM

Except that the problem says
(Use the average temperature for the entropy calculation.)

These vague statements like "I've tried everything and can't get the answer" are pretty useless for figuring out where the difficulty lies.

OlderDan said:
Except that the problem says "average temperature"

These vague statements like "I've tried everything and can't get the answer" are pretty useless for figuring out where the difficulty lies.
Ok. Then the problem Jayhawk must be having is in the determination of average temperatures for the cool and warm water. Jayhawk, what are you using for average temperatures?

AM

## What is entropy change?

Entropy change is a measure of the amount of disorder or randomness in a system. It is commonly denoted as ΔS and is measured in joules per kelvin (J/K).

## How is entropy change calculated?

Entropy change is calculated using the formula ΔS = Q/T, where Q is the heat transferred and T is the temperature in kelvin.

## What is the formula for calculating entropy change for a phase change?

The formula for calculating entropy change during a phase change is ΔS = ΔH/T, where ΔH is the enthalpy change and T is the temperature in kelvin.

## Can entropy change be negative?

Yes, entropy change can be negative. This indicates a decrease in disorder or randomness in a system.

## How does the mass and temperature affect the entropy change?

The mass and temperature of a substance can affect the entropy change. In general, as the mass and temperature increase, so does the entropy change. This is because there is more energy and movement in the system, resulting in increased disorder and randomness.

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