## 4A.9

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 4A.9

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A piece of copper of mass 20.0 g at 100.0 C is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 C. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings.

In the solutions manual, the equation to solve for the final temperature is given by heat lost by metal = –heat gained by water. This gives q = –q, and so mC$\Delta$T = –mC$\Delta$T. Once you substitute everything in, all of the variables are known except for Tfinal (final temperature) on both sides of the equation. So, the next step should be to isolate both sides with Tfinal so that you can find what its value is. But how do you know that the final temperature is the same when the metal loses heat and the water gains heat?

J_CHEN 4I
Posts: 54
Joined: Tue Nov 14, 2017 3:01 am

### Re: 4A.9

Plug the final temperature into the equation and if mCdeltaT = -mCdeltaT is true, then the final temperatures are the same.

GFolk_1D
Posts: 101
Joined: Fri Aug 09, 2019 12:15 am

### Re: 4A.9

Also it says that no heat is lost to surroundings so we can assume that heat moves out of the copper and into the water, eventually making their temperatures equal to one another.

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