Find ΔE
a. when 175 J of work is done on a system that evolves 50 J of heat to the surroundings
b. when a gas absorbs 28 kJ of heat and has 13 kJ work done on it
Solution.
a)ΔE=A−Q;a)\Delta\Epsilon=A-Q;a)ΔE=A−Q;
ΔE=175J−50J=125J;\Delta E=175J-50J=125J;ΔE=175J−50J=125J;
b)ΔE=Q−A;b)\Delta E=Q-A;b)ΔE=Q−A;
ΔE=28kJ−13kJ=15kJ;\Delta E=28kJ-13kJ=15kJ;ΔE=28kJ−13kJ=15kJ;
Answer: a)ΔE=125J;a)\Delta E=125J;a)ΔE=125J;
b)ΔE=15kJ.b)\Delta E=15kJ.b)ΔE=15kJ.
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