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### Homework 4A.9

Posted: Wed Jan 29, 2020 8:38 pm
For homework problem 4A.9:
A piece of copper of mass 20.0 g at 100.0 8C is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 8C. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings.

The solutions manual states that we solve for the final temperature by equaling the heat lost by the metal to the heat gained by the water. However, it also states that the heat gained by the water is negative. Why is it negative in this case? Doesn't a negative q mean that the water is losing heat (exothermic)? Why isn't the heat lost by the metal a negative value?

### Re: Homework 4A.9

Posted: Wed Jan 29, 2020 8:57 pm
I think we went over something similar during lecture where Dr. Lavelle said the heat released by a reaction = - heat absorbed by the solution. The negative sign is just referring to the heat being released, which is negative. Heat released = - (heat absorbed), while - (heat released (-)) = heat absorbed.

### Re: Homework 4A.9

Posted: Fri Jan 31, 2020 9:34 pm
I think that your reasoning is exactly correct. And I believe that it doesn't matter which side you would like to put the negative sign on. The only thing matter is that one substance is losing heat whereas the other one is gaining heat, so if we set the change in heat of one substance negative, then the other is positive, and vice versa.

### Re: Homework 4A.9

Posted: Sun Feb 02, 2020 1:33 pm
The negative sign is there to help signify the relationship between the two temperatures and it doesn't really matter because you just want to have the right amounts of negatives on both sides.