A system does 210 J of work on its environment and gains 80.6 J of heat in the process. Find the change in the internal energy of (a) the system and (b) the environment.

I am not sure how to start this.

ΔU = Q - W

Where ΔU is the change in the internal energy of the system,
Q is the heat/energy gained from the surroundings,
W is the work done on the environment.

ΔU =80.6-210=-129.4 J.
The internal energy decreased by 129.4 Joules which corresponds to a temperature decrease of the system

To find the change in the internal energy of the system and the environment, we can apply the first law of thermodynamics, also known as the Law of Conservation of Energy. According to this law, the change in internal energy of a system is equal to the work done on the system (W) plus the heat added to the system (Q).

(a) Change in internal energy of the system:
ΔU_system = W + Q

Given:
Work done on the system (W) = 210 J
Heat added to the system (Q) = 80.6 J

Substituting the values into the equation:
ΔU_system = 210 J + 80.6 J
ΔU_system = 290.6 J

Therefore, the change in internal energy of the system is 290.6 J.

(b) Change in internal energy of the environment:
Since the system is doing work on the environment, the work is done by the system on the surroundings. By the sign convention, work done by the system is negative.

W = -210 J (negative because work is done on the environment)

Using the first law of thermodynamics:
ΔU_environment = W + Q

Given:
Work done on the environment (W) = -210 J
Heat added to the system (Q) = 80.6 J

Substituting the values into the equation:
ΔU_environment = -210 J + 80.6 J
ΔU_environment = -129.4 J

Therefore, the change in internal energy of the environment is -129.4 J (negative because work is done by the system on the environment).