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### Problem 4A.9) 7th edition of the textbook

Posted: **Wed Jan 30, 2019 2:23 pm**

by **Katie Frei 1L**

For problem 4A.9 of the 7th edition of the textbook, I got the correct final temperature to be 25 degrees Celsius. However, when I plugged this value back into the equation in which I solved for final temperature, the value was not checking out. I was wondering, is there a way in which to check in these situations that the final temperature is correct? Thanks!

### Re: Problem 4A.9) 7th edition of the textbook [ENDORSED]

Posted: **Thu Jan 31, 2019 10:58 am**

by **Chem_Mod**

This is because of significant figures.

It will work when plugging in 25 °C. The -∆T for copper is 75 °C and the ∆T for water is 3 °C (notice this now has one sig fig). When you multiply by the masses and the heat capacities, you get 570 J for copper and 636.3864 J for water, which are both 600 J to one sig fig.

There is another order in which you can do the operations which gives you 24.7 °C as the correct number of sig figs. This gives you 75.3 °C for the -∆T of copper and 2.7 °C for the ∆T of water, which translate to 572.28 J and 572.74776 J. These are both 570 J to two sig figs.

There is another method for determining uncertainties instead of using significant figures which is more rigorous but beyond the scope of this class.

### Re: Problem 4A.9) 7th edition of the textbook

Posted: **Wed Feb 06, 2019 8:22 pm**

by **bonnie_schmitz_1F**

In this problem, the solutions manual says that heat lost by metal = - heat gained by water. Why would the heat gained by water be negative?

### Re: Problem 4A.9) 7th edition of the textbook

Posted: **Thu Feb 07, 2019 12:07 am**

by **paytonm1H**

responding to the question on heat lost by ice = - heat gained by water...

The hear gained by the water is not negative. This relationship is simply implying that the heat lost by the ice and the heat gained by the water are equal and opposite one another. So you can use that relationship to find the final temperature.

I hope this helped

### Re: Problem 4A.9) 7th edition of the textbook

Posted: **Thu Feb 28, 2019 4:53 am**

by **Kobe_Wright**

bonnie_schmitz_1F wrote:In this problem, the solutions manual says that heat lost by metal = - heat gained by water. Why would the heat gained by water be negative?

Well thinking about it, if the metal loses heat its value will be negative and since that is negative, H(metal)=-H(water), H(metal)=-x, H(water)=-(-H(metal)= some + value.

### Re: Problem 4A.9) 7th edition of the textbook

Posted: **Wed Mar 13, 2019 1:04 pm**

by **bonnie_schmitz_1F**

Thank you!

### Re: Problem 4A.9) 7th edition of the textbook

Posted: **Mon Oct 28, 2019 12:44 am**

by **RichBollini3C**

Does anyone know what sections we can turn in for homework number 5?

### Re: Problem 4A.9) 7th edition of the textbook

Posted: **Sun Nov 17, 2019 9:53 pm**

by **bloodorangefield**

When energy is absorbed by water, does it release heat?