A system with an increase in internal energy of 80 J and at the same time has 50 J of work done on it by the surrounding. What is the heat change of the system?
The first law of thermodynamics states that the change in internal energy of a system equals the heat added to the system minus the work done by the system:
ΔU = Q - W
where:
ΔU = change in internal energy
Q = heat added to the system
W = work done by the system
In this case, we have:
ΔU = 80 J
W = -50 J (since work is being done on the system by the surroundings, the sign is negative)
So, we can rearrange the equation to solve for Q:
Q = ΔU + W
Q = 80 J + (-50 J)
Q = 30 J
Therefore, the heat change of the system is 30 J.
To find the heat change of the system, you can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):
ΔU = Q - W
Given that the internal energy change (ΔU) is 80 J and the work done on the system (W) is 50 J, we can substitute these values into the equation:
80 J = Q - 50 J
To solve for Q, we can rearrange the equation:
Q = 80 J + 50 J
Q = 130 J
Therefore, the heat change of the system is 130 J.