A 100 w lightbulb is place in a cylinder equipped with a moveable piston. The ligshtbulb is turned on for 0.015 hours, and the assembly expands from initial volume of 0.85 L to a final volume of 5.88 L against an external pressure of 1.0 atm. Use the wattage of the lightbulb and the time it is on to calculate delta E in joules (assume that the cylinder and lightbulb assembly is the system and assume two significant figures). calculate w and q..
work = -pdV = -p(Vfinal - Vinitial)
Convert 100 w to watt*sec = joules and that will be q
dE = q+w
Well, well, well, let's shed some light on this question and have some fun with it!
First, let's calculate the energy change (ΔE) using the formula ΔE = PΔV, where P is the external pressure and ΔV is the change in volume. Plugging in the values, we have:
ΔE = 1.0 atm * (5.88 L - 0.85 L) = 4.986 L·atm
But this is in liters of atmosphere, not joules. So, we need to convert it to joules by using the conversion factor 1 L·atm = 101.3 J:
ΔE = 4.986 L·atm * 101.3 J / 1 L·atm = 504.7298 J ≈ 505 J
Alrighty, now let's move on to work (w) and heat transfer (q).
To calculate work, we can use the formula w = -ΔE since the volume expansion is against an external pressure. Thus:
w = -505 J
Now, for heat transfer (q), we can use the formula q = ΔE + w. Plugging in the values:
q = 505 J + (-505 J) = 0 J
That's right, folks! The heat transfer (q) for this scenario is zero. Howe-hilarious is that? So, no extra energy was transferred as heat.
In summary:
- ΔE = 505 J (energy change)
- w = -505 J (work)
- q = 0 J (heat transfer)
I hope that cleared things up and brought some laughter to your day!
To calculate the change in energy (ΔE), work (w) and heat (q) for this system, we can use the first law of thermodynamics, which states that the change in energy of a system is equal to the heat added to the system minus the work done by the system.
Since the lightbulb is turned on for 0.015 hours, the amount of energy it emits can be calculated using the equation:
Energy (E) = Power (P) x Time (t)
Given that the power of the lightbulb is 100 watts and the time it is on is 0.015 hours:
E = 100 W x 0.015 h
E = 1.5 watt-hours
To convert watt-hours to joules, we need to multiply by the conversion factor 3,600 J/Watt-hour:
E = 1.5 watt-hours x 3,600 J/Watt-hour
E = 5400 J
So, the energy emitted by the lightbulb is 5400 joules.
Now, let's calculate the work done by the system using the equation:
Work (w) = -Pext x ΔV
Where Pext is the external pressure and ΔV is the change in volume of the system.
Given that the external pressure is 1.0 atm and the change in volume is from 0.85 L to 5.88 L:
ΔV = (5.88 L) - (0.85 L)
ΔV = 5.03 L
Converting liters to joules, we use the conversion factor 101.3 J/L-atm:
w = -(1.0 atm) x (5.03 L) x (101.3 J/L-atm)
w = -509.349 J
Since the work done by the system is negative (indicating work done on the system), we'll take the absolute value:
w = 509.349 J
Finally, let's calculate heat (q) by subtracting the work from the energy emitted:
q = ΔE - w
q = 5400 J - 509.349 J
q = 4890.651 J
Therefore, the change in energy (ΔE) is 5400 joules, the work done (w) is 509.349 joules, and the heat exchanged (q) is 4890.651 joules.
w=500
q=291
dE = 791