Exercise 8.19, proper use of equations?

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Eric Lam 2K
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Joined: Wed Sep 21, 2016 2:58 pm

Exercise 8.19, proper use of equations?

Postby Eric Lam 2K » Sat Jan 21, 2017 1:38 pm

Exercise 8.19 is follows:

(a) Calculate the heat that must be supplied to a 500.0g copper kettle containing 400.0g of water to raise its temperature from 22 degrees Celsius to the boiling point of water, 100 degrees Celsius.
(b) What percentage of the heat is used to raise the temperature of the water?

I'm using the equation q=(n) (Cv,m) (delta T)

where n = mols of Cu and H2O ( 7.868 mol and 22.2 mol respectively )
Cv,m = the molar heat capacity of copper and water (24.4 J/K mol and 75.2 J/K mol respectively)
delta T = change in temperature in Kelvin ( total change is 78 K )

My understanding is that I calculate q(Cu) and q(H2O) with their respective n and Cv,m and use delta T = 78 K, add the products of the equation and that is my answer for (a). Then, for (b), I simply divide q(Cu) by q(H2O) to find the percentage of the heat that is used to raise the temperature of the water.

My question is, did I properly pick the Cv,m value and did I pick the right equation for this problem, and why instead of using the equation [ deltaH = deltaU + n R deltaT ] I use [ q=(n) (Cv,m) (delta T) ] ? And will Cv,m values be given on the quiz/exams?

Danny Nguyen 2H
Posts: 17
Joined: Fri Jul 22, 2016 3:00 am

Re: Exercise 8.19, proper use of equations?

Postby Danny Nguyen 2H » Sat Jan 21, 2017 2:45 pm

I think using the equation for q=(n) (Cv,m) (delta T) is the best for this question because it's asking to find the change in heat and the equation does just that. For the C values I'm pretty sure we will have to be given those values to solve the problem using the equation above. And [ deltaH = deltaU + n R deltaT ], can't be applied because for this particular problem internal energy is not given. In addition I believe you derived the equation wrong. If you wanted to solve for deltaH the equation would actually have to be [ deltaH = deltaU + delta n R T ]. Its the change in moles not temperature.

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