Homework Problem 8.19

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Mike Matthews 1D
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Joined: Fri Sep 29, 2017 7:04 am

Homework Problem 8.19

Postby Mike Matthews 1D » Thu Jan 11, 2018 2:14 pm

Homework problem 8.19 reads, "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 the boiling point of water, 100 degrees Celsius. What percentage of the heat is used to raise the temperature of the water?" Given that copper and water have different heat capacities how would you go about doing this problem?

Jared Smith 1E
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

Re: Homework Problem 8.19

Postby Jared Smith 1E » Thu Jan 11, 2018 2:28 pm

For this problem, you would calculate the q values for raising the copper to 100 degrees and the water to 100 degrees separately, and then add the q values together to get the final answer.

Nishma Chakraborty 1J
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

Re: Homework Problem 8.19

Postby Nishma Chakraborty 1J » Thu Jan 11, 2018 2:36 pm

Hey!

For part a) of this problem, I used the equation used the equation q=m*Cs*change in T. I converted the mass into kilograms and used Table 8.2 to find the specific heat capacity (Cs) of both water and copper. So, for water I had: 0.4kg*4.184J*78=130.5kJ, and for copper I had: 0.5kg*0.38J*78=14.82J. Total heat supplied equals qwater + qcopper, so 145.32kJ.

For part b) I just used ((amount of heat supplied for water)/(total heat supplied))*100, and got an answer of 89.8%.

Hope this helps! :)

Wilson Yeh 1L
Posts: 42
Joined: Fri Sep 29, 2017 7:06 am

Re: Homework Problem 8.19

Postby Wilson Yeh 1L » Thu Jan 11, 2018 7:25 pm

I did what the above posters did as well and got the same answer. I'm just curious as to why the book answer is 1.4 x 10^2 kJ? Shouldn't it be 1.5 if anything?


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