Chapter 8 Number 19

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Vanessa Romero-Campos 2B
Posts: 39
Joined: Fri Sep 25, 2015 3:00 am

Chapter 8 Number 19

Postby Vanessa Romero-Campos 2B » Sun Jan 15, 2017 5:12 pm

a)calculate the heat that must be supplied to a 500.0 g copper kettle containing 400.0 g of water to raise its temperature from 22.0 degrees Celsius to 100degrees celsius.

I used the equation q=(m)*(Cs)*(delta T) and plugged in the given information as follows:

(400g)*(4.184 J/C*g)*(100-22) and got 13040.8, however, on the solution manual there is a second part that is added to the part I already have.

Why is it that the second part ((500g)*(0.38 J/C*g)*(100-22)) is added to the first?

The solution manual has the following solution:
(400g)*(4.184 J/C*g)*(100-22)+(500g)*(0.38 J/C*g)*(100-22)

Gisselle Cervantes 2G
Posts: 17
Joined: Wed Sep 21, 2016 2:58 pm

Re: Chapter 8 Number 19

Postby Gisselle Cervantes 2G » Sun Jan 15, 2017 5:20 pm

What you calculated was only the first part that raises the temperature of the copper, so the second part was added to take into account the heat needed to raise the temperature of the water. When calculating heat change, you need to take into account both the metal and water.

Vanessa Romero-Campos 2B
Posts: 39
Joined: Fri Sep 25, 2015 3:00 am

Re: Chapter 8 Number 19

Postby Vanessa Romero-Campos 2B » Sun Jan 15, 2017 5:22 pm

Ohh okay, Thank you :)


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