1. Water with a mass of 2kg is held at constant volume in a pressure cooker. 10,000 J of energy is added from the heater inside the pressure cooker. The container is not terribly well insulated, so the container leaks 2000 J of energy to the surrounding air. What is the total change in the systems internal energy?
here are the equations we can use:
(-m1)(C1)(ΔT1)=(m2)(C2)(ΔT2)
Q=mcΔT
Q=mL
ΔU=Q-W
W=PΔV
There might be others I am forgetting...sorry
Please indicate which equation you use and why you used it. Thanks!
changeU=Qadded-loss(W)
changeU=10,000-2000 J
To calculate the total change in the system's internal energy, we will use the equation ΔU = Q - W.
Here, ΔU represents the change in internal energy, Q represents the heat added to the system, and W represents the work done by the system.
In this scenario, the volume is held constant, so there is no work done (W = 0).
The heat added to the system (Q) can be calculated using the equation Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The heat lost to the surroundings (Leakage) can be subtracted from the total heat added to the system.
Hence, the equation ΔU = Q - Leakage can be applied here.
To find the total change in the system's internal energy, we can use the equation ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done on or by the system.
In this case, we are given that 10,000 J of energy is added from the heater inside the pressure cooker (Q = 10,000 J) and 2000 J of energy is leaked to the surrounding air. The system is held at constant volume, so there is no work done (W = 0).
Using the equation ΔU = Q - W, we can substitute the values we have:
ΔU = Q - W
ΔU = 10,000 J - 0 J
ΔU = 10,000 J
Therefore, the total change in the system's internal energy is 10,000 J.
Note: We did not use the other equations you mentioned because they are used in different contexts and not applicable to this specific scenario.