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.

Q = - 730 KJ

730 KJ is transferred out of the system

Explanation:

According to the first law of thermodynamics, energy can neither be created nor destroyed, but can be transformed from one form to another.

For a particular process/system, the first law is interpreted as

ΔU = Q + W (depending on convention, some textbooks give this relation as ΔU = Q - W)

ΔU is the change in internal energy of the system, in my convention, it is positive if the internal energy increases and negative otherwise.

Q = Heat transferred into or out of the system. Q is positive for heat transferred into the system and negative for heat transferred out of the system.

W = workdone by the system or work done on the system. W is positive for workdone on the system and W is negative for when work is done by the system.

ΔU decreases by 250 KJ, that is, ΔU = - 250 KJ

Q = ?

W = 480 KJ (Work is done on the system)

- 250 = Q + 480

Q = - 250 - 480 = - 730 KJ

The heat is transferred out of the system.

protons in

it becomes positively charged by charging it