Answer:
a) 7.2 × 10³ J
b) -5.1 × 10² J
c) 7.7 × 10³ J
Explanation:
a)Use the wattage of the light bulb and the time it is on to calculate ΔU in joules (assume that the cylinder and light bulb assembly is the system and assume two significant figures). Express your answer using two significant figures.
The light bulb has a power of 100 W (100 J/s) and works during a time of 2.0 × 10⁻² hours. The change in the internal energy (ΔU) is:
[tex]\Delta U=2.0\times 10^{-2} h.\frac{3600s}{1h} .\frac{100J}{s} =7.2 \times 10^{3} J[/tex]
b) Calculate w. Express your answer using two significant figures.
We can calculate the work (w) using the following expression.
w = -P . ΔV
where,
P is the external pressure
ΔV is the change in the volume
[tex]w=-1.0atm\times (5.88L-0.85L).\frac{101.325J}{1atm.L} =-5.1 \times 10^{2} J[/tex]
c) Calculate q. Express your answer using two significant figures.
We can calculate the heat (q) using the following expression.
ΔU = q + w
q = ΔU - w = 7.2 × 10³ J - (-5.1 × 10² J) = 7.7 × 10³ J