Homework Problem 8.19 and 8.21

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Maggie Elgin 2A
Posts: 19
Joined: Wed Sep 21, 2016 2:57 pm

Homework Problem 8.19 and 8.21

Postby Maggie Elgin 2A » Sun Jan 22, 2017 7:42 pm

Could someone please explain to me how to do these problems?

Khachik_Hmayakyan_2E
Posts: 25
Joined: Sat Jul 09, 2016 3:00 am

Re: Homework Problem 8.19 and 8.21

Postby Khachik_Hmayakyan_2E » Sun Jan 22, 2017 8:48 pm

In order for the heat to be transferred to the water, it must first raise the temperature of the copper. In both cases, we use q=mC[delta]T where we input the mass of the copper kettle (given), the specific heat capacity of copper (table 8.2), and the change in temperature (final T -initia T). Repeat this same process for the values given for water and add them to give the heat required to raise the temperature. To calculate the percentage you simply place the amount of heat required to raise the temp of water over the total heat required to raise the temp overall and multiply by 100. Q water/Q total • 100

Josh Ku 3H
Posts: 25
Joined: Wed Sep 21, 2016 2:59 pm

Re: Homework Problem 8.19 and 8.21

Postby Josh Ku 3H » Sun Jan 22, 2017 9:32 pm

For #21,

The question gives you the starting temperature of both the copper and the water as well as the mass for both of them. What you want to calculate is the final temperature. Since the copper is in the water, the two should have the same final temperature. With this in mind you use the equation: q = mc(delta t). In addition, the question also says "Assume that no energy is lost to the surroundings". This means that the heat lost from the copper is equal to the heat gained by the water. Because of this, we can set the two specific heat equations equal to each other but since one is losing heat while the other is gaining, you have make one of the equations negative in order for them to be equivalent. From then, you just solve for the final temperature.

Hope this helps!


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