Answer:
ΔU = - 310.6 J (negative sign indicates decrease in internal energy)
W = 810.6 J
Explanation:
a.
Using first law of thermodynamics:
Q = ΔU + W
where,
Q = Heat Absorbed = 500 J
ΔU = Change in Internal Energy of Gas = ?
W = Work Done = PΔV =
P = Pressure = 2 atm = 202650 Pa
ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³
Therefore,
Q = ΔU + PΔV
500 J = ΔU + (202650 Pa)(0.004 m³)
ΔU = 500 J - 810.6 J
<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>
<u></u>
b.
The work done can be simply calculated as:
W = PΔV
W = (202650 Pa)(0.004 m³)
<u>W = 810.6 J</u>
Answer:
.067 so C
Explanation:
I asked my sister who is in 2nd grade and she said it was right so you are good! =). have a great day!
What is the heat extracted from the cold reservoir for the refrigerator shown in(Figure 1) ? Assume that W1 = -123J and W2 = 88J .
<span>Qc= _________ </span>
<span>Part B
</span>
K=105J
