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. Thank you for the help.
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 internal energy of the system, you can use the first law of thermodynamics, which states that the change in internal energy (∆U) of a system is equal to the heat flow into the system (Q) minus the work done by the system on its surroundings (W). Mathematically, it can be represented as:
∆U = Q - W
Given that the system does 210 J of work on its environment (W = -210 J) and gains 80.6 J of heat (Q = 80.6 J), we can substitute these values into the equation to find the change in internal energy of the system.
(a) Change in internal energy of the system:
∆U = Q - W
∆U = 80.6 J - (-210 J)
∆U = 80.6 J + 210 J
∆U ≈ 290.6 J
So, the change in internal energy of the system is approximately 290.6 J.
To find the change in internal energy of the environment, we can use the fact that the total change in energy in an isolated system is zero. In other words, the change in internal energy of the system plus the change in internal energy of the environment should equal zero.
(b) Change in internal energy of the environment:
∆U_environment = -∆U_system
∆U_environment = -290.6 J
Therefore, the change in internal energy of the environment is approximately -290.6 J (negative indicates a decrease in internal energy).
Please note that the values provided are assumed to be positive, indicating a gain or increase in energy. If the signs of work and heat were provided differently in the problem statement, the results may vary accordingly.