I'm stuck on part B of this question:

A system does 125 J of work on environment and gains 79 J of heat in the process. Find change in internal energy (delta U, abbreviated dU)of
a. the system and b. the environment.

A. Find dU of the system
W = 125 J
Q = 79 J
dU = Q - W
dU = (79 J - 125 J) = -46.0 J

B. find dU of the environment

To find the change in internal energy (dU) of the environment, we can use the fact that the total change in internal energy of a system and its surroundings is always zero in an isolated system. This is known as the First Law of Thermodynamics or the Law of Conservation of Energy.

Since we already know the change in internal energy of the system is -46 J (as calculated in part A), the change in internal energy of the environment can be found by simply taking the negative of that value.

Therefore, the change in internal energy of the environment (dU_environment) would be:

dU_environment = -(-46 J)
dU_environment = 46 J

So, the change in internal energy of the environment is positive 46 J.