1. This is actually a homework question...

A 100g iron bar at 100 celcius is placed in 200g of water at 20 celsius. If the heat capacity of iron is 0.11 cal/g celcius to what final temperature will the iron bar cool?

My teacher gave me this equation Q=mC(delta)T or heat transferred=massXheat capacityXChange in temperature.

My problem is, is I am not sure if I am supposed to solve for delta T and if I do I have no way of finding out heat transfered.

2.

3. I'm going to say.....40C. Since if either were the same mass it would be 60 evened out, but since there is 2x as much water, the difference 40 is halved into 20, so that the outcome is 40 (20 + 20). I don't think you even need to use the equation; the entropy of the iron should be linear and predictable. Just a guess based on logic and no math. If I'm right, I'll feel allot more confident about by logical guessing skills.

4. The heat lost by the iron bar is equal to the heat gained by the water. The final temperature, T, of each will be the same. The delta T for the iron will be 100 minus the final temperture. The delta T for the water will be the final temperature minus 20. The heat capacity of water is 1 calorie per gram-degree C. Knowing these facts, you should be able to set up an equation and solve for the final temperature T.

5. I'm going to say.....40C. Since if either were the same mass it would be 60 evened out, but since there is 2x as much water, the difference 40 is halved into 20, so that the outcome is 40 (20 + 20). I don't think you even need to use the equation; the entropy of the iron should be linear and predictable. Just a guess based on logic and no math. Very Happy If I'm right, I'll feel allot more confident about by logical guessing skills. Smile
You have to take in to acount the heat capacity of the two substances.

(mass of iron)(final iron temp - initial iron temp)(specific heat of iron) = (mass of H2O)(initial H2O temp - final H2O temp)(specific heat of H2O)

6. Well how do you find the final temp.

Like this?

100-x=x-20

120=2x

60=x

7. Originally Posted by EV33
Well how do you find the final temp.

Like this?

100-x=x-20

120=2x

60=x
No, you have to use the equation for Q that your teacher gave you.

8. Is it like this...

(100g)(.11cal/gC)(100C-x)=(200g)(1cal/gC)(x-20C)

(11Cal/C)(100C-x)=(200Cal/C)(x-20C)

1100Cal-(11Cal/C)x=200Cal/C-4000Cal

5100Cal=(211Cal/C)X

5100Cal/211Cal/C=x

24.171 Celcius=x

9. Looks good to me.

10. Thank you all for the help

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