The figure shows a pV diagram of a gas for a complete cycle. During part bc of the cycle, 1190 J of heat flows into a system, and at the same time the system expands against a constant external pressure of 7.00 × 104 Pa as its volume increases from 0.0200 m3 to 0.0800 m3. Calculate the change in internal (thermal) energy of the system during part bc of the cycle. If the change is nonzero, be sure to indicate whether the change is positive or negative.

Please show work

1050j

To calculate the change in internal (thermal) energy of the system during part bc of the cycle, we can use the first law of thermodynamics:

∆U = Q - W

Where:
∆U is the change in internal energy of the system
Q is the heat added to the system
W is the work done by the system

In this case, we are given that 1190 J of heat flows into the system, which means Q = 1190 J.

The work done by the system can be calculated using the formula:

W = -P∆V

Where:
P is the constant external pressure
∆V is the change in volume

In this case, the constant external pressure, P, is given as 7.00 × 10^4 Pa. The change in volume, ∆V, is the final volume minus the initial volume, which is (0.0800 m^3 - 0.0200 m^3) = 0.0600 m^3.

Plugging in the values into the formula, we get:

W = - (7.00 × 10^4 Pa) * (0.0600 m^3)
W = - 4200 J

Therefore, the work done by the system is -4200 J.

Now, we can calculate the change in internal energy using the first law of thermodynamics:

∆U = Q - W
∆U = 1190 J - (-4200 J)
∆U = 1190 J + 4200 J
∆U = 5390 J

Therefore, the change in internal energy of the system during part bc of the cycle is 5390 J. Since the change in internal energy is positive, it means the internal energy of the system has increased during this part of the cycle.