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Nzy2GU1AFkhM_1210016470_jwk572

# Nzy2GU1AFkhM_1210016470_jwk572 - homework 14 KIM JI Due Dec...

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homework 14 – KIM, JI – Due: Dec 5 2007, 4:00 am 1 Question 1, chap 21, sect 1. part 1 of 3 10 points Determine the change in the internal energy of a system that absorbs 322 cal of thermal energy while doing 640 J of external work. Correct answer: 707 . 892 J (tolerance ± 1 %). Explanation: Let : Q = 322 cal and W = 640 J . According to the first law of thermodynam- ics, we have Δ U = Q - W , where Q is the thermal energy transferred into the system and W is the work done by the system. Then we have Δ U = (322 cal) 4 . 186 J cal - 640 J = 707 . 892 J . Question 2, chap 21, sect 1. part 2 of 3 10 points Determine the change in the internal energy of a system that absorbs 202 cal of thermal energy while 586 J of external work is done on the system. Correct answer: 1431 . 57 J (tolerance ± 1 %). Explanation: Let : Q = 202 cal and W = - 586 J . Δ U = Q - W = (202 cal) 4 . 186 J cal + 586 J = 1431 . 57 J . Question 3, chap 21, sect 1. part 3 of 3 10 points Determine the change in the internal energy of a system that is maintained at a constant volume while 1020 cal is removed from the system.

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Nzy2GU1AFkhM_1210016470_jwk572 - homework 14 KIM JI Due Dec...

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