## hmwrk 8.21 6th edition

Lina Petrossian 1D
Posts: 77
Joined: Fri Oct 05, 2018 12:16 am

### hmwrk 8.21 6th edition

8.21 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.

Why do we use this setup:

20 (mass)*0.39 (specific heat of Cu) *(100-T) (change in temperature) = 50.7 (mass)* 1 (specific heat of H2O) *(T-22)(change in temperature)

ryanhon2H
Posts: 60
Joined: Fri Sep 28, 2018 12:28 am

### Re: hmwrk 8.21 6th edition

The copper is losing heat to the water, so the equations would be set equal to each other. However, you would have to flip the sign of the copper because the heat lost would be negative, so you get

-qcopper = qwater

By using (100-T) you get rid of the negative sign so both values become positive. You can use (T-100) but then the equation would look like

- (20 (mass)*0.39 (specific heat of Cu) *(T-100) (change in temperature)) = 50.7 (mass)* 1 (specific heat of H2O) *(T-22)(change in temperature)

with that negative sign in the front.

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