# Heat transfer

1. May 9, 2005

### Msnyder

First time here so be easy please!!

Problem: What would final temp. be when 100g of 25C water is mixed with 75g of 40c water?

Do I use deltaT=deltaQ/cm?

If there is a more basic forum, please let me know, I do not want to offend anyone.

Last edited: May 9, 2005
2. May 9, 2005

### Galileo

Yep, that's the equation to use.

3. May 9, 2005

### Msnyder

any help would be appreciated, I am getting confused with the problem

somehow I am getting 15C/175,000????

By the way, if there is a more basic forum, let me know....I do not want to offend anyone!

Last edited: May 9, 2005
4. May 9, 2005

### inha

what did you do to get that answer?

the heat released by the hotter portion is the heat gained by the cooler portion. that's where I'd start solving the problem from.

addition: maybe this should be in some of the homework forums? otherwise it's in the right place. if this is in the wrong place don't start a new thread but wait for a moderator to move it to the right place.

5. May 9, 2005

### Msnyder

I understand that heat lost is = to heat gained,but what about the added water?

I guess the better question would be, what is the c of water?

**I feel like a 10 yr old boy in this class!

Last edited: May 9, 2005
6. May 9, 2005

### inha

c is 4.19kJ/kg*K.

Think in terms of conservation of energy. If no energy is lost then the Q of the final state (175 grams of water at some temperature T) must be equal to... You fill the blank.

7. May 9, 2005

### Msnyder

Quick questions, do I average the temps or do I subtract 40 -25 which is 15C. Are you saying that the water conserves it's energy, so the addition of mass does not matter, it's just 15C?

8. May 9, 2005

### Galileo

Don't be confused by the fact that you're mixing water with water. In the end (at thermal equilibrium) both 'parts' share the same temperature. Call it T.
Using this T, what is the amount of heat 1 part of the water gained, what is the amount of heat the other one gained? What should T be such that the total amount of heat gained is zero?

9. May 9, 2005

### Msnyder

wow....I guess one gained 15 degrees and the other gained nothing, so together they gain 15, because the transfer of heat is from the hotter to the cooler?????

So, from the original problem, the water ends up at 40 degrees?

Last edited: May 9, 2005
10. May 9, 2005

### Msnyder

Thanks for your help everyone, but I am not getting anywhere. This is just extra hard for me at this point. I will try my text again and see if I can come up with something. Thanks again,
Mike

Last edited: May 9, 2005
11. May 9, 2005

### Andrew Mason

Think of the problem this way:

Begin with 175 g of water at 25 deg. C and add heat that is sufficient to warm 75g of water to 40 degrees (ie. 75 g x 15 deg = 1125 Cal). What is the temperature raised by? (ie. 1125 Cal into 175 g = ? deg).

AM