In a certain process, the energy of the system decreases by 250 kJ. The process involves 480 kJ of work done on the system. Find the amount of heat Q transferred in this process. Express your answer numerically in kilojoules. Make your answer positive if the heat is transferred into the system; make it negative if the heat is transferred into the surroundings.
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In this case, we are given that ΔE = -250 kJ (the energy decreases, so the change is negative) and W = 480 kJ. We are asked to find Q. Substituting the given values into the equation, we get -250 kJ = Q - 480 kJ. Solving for Q, we find that Q = -250 kJ + 480 kJ Show more…
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In a certain process, $8.00$ kcal of heat is furnished to the system while the system does $6.00 \mathrm{~kJ}$ of work. By how much does the internal energy of the system change during the process? Here $8.00$ kcal is heat-in and $6.00 \mathrm{~kJ}$ is work-out, both of which are positive. Consequently, $\Delta Q=(8000$ cal $)(4.184 \mathrm{~J} / \mathrm{cal})=33.5 \mathrm{~kJ}$ and $\Delta W=6.00 \mathrm{~kJ}$ Therefore, from the First Law $\Delta Q=\Delta U+\Delta W$, $$\Delta U=\Delta Q-\Delta W=33.5 \mathrm{~kJ}-6.00 \mathrm{~kJ}=27.5 \mathrm{~kJ}$$
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