7th Edition, 4A.7

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Andrew Bennecke
Posts: 62
Joined: Fri Sep 28, 2018 12:15 am

7th Edition, 4A.7

Postby Andrew Bennecke » Sun Feb 03, 2019 3:03 pm

I think I am missing a step somewhere for 4A.7 (part a) in the 7th edition because I'm plugging in the given values to the equation, but can't obtain the correct answer.
q=m*C*(Delta T)
q=(400g)(0.38 J*C^-1*g^-1)(80 C-1)
q=12.2 kJ
The textbook says that the answer is 112 kJ, but I can't figure out where I messed up.

Lexie Baughman 2C
Posts: 30
Joined: Sat Oct 06, 2018 12:16 am

Re: 7th Edition, 4A.7

Postby Lexie Baughman 2C » Sun Feb 03, 2019 4:48 pm

The heat change will be made up of two terms - one term to raise the temperature of the copper and the other to raise the temperature of the water.
q = (300g)(4.18J/Cg)(100C-20C) + (400g)(.38J/Cg)(100C-20C) = 1.12x10^5 J = 1.12x10^2 kJ (with C being Celsius)

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