If a system has 2.50 x 10 kcal of work done to it, and releases 5.00...

Question

Question

Answer 1

Answer #1

W = 2.50*10^2 kcal

= 2.50*10^2 * 4.184 kJ

= 1.05*10^3 kJ

Q = -5.00*10^2 kJ

Negative because heat has been released

Use:

Delta E = Q+W

= -5.00*10^2 kJ + 1.05*10^3 kJ

= 5.50*10^2 kJ

Answer: 5.50*10^2 kJ

Answer 2

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