Capacitor A and capacitor B both have the same voltage across their plates. However, the energy of capacitor A can melt m kilograms of ice at 0 °C, while the energy of capacitor B can boil away the same amount of water at 100 °C. The capacitance of capacitor A is 8.4 μF. What is the capacitance of capacitor B?
Please note that for water, the latent heat of fusion is Lf = 33.5 x 104 J/kg, and the latent heat of vaporization is Lv = 22.6 x 105 J/kg
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What is the Capacitance of capicitor b?
Q₁=Lf•m = C₁V²/2
Q₂=Lv•m = C₂V²/2
Lf•m/ Lv•m=2C₁V²/2 C₂V²
C₂=Lv•C₁/Lf = 22.6•10⁵•8/33.5•10⁴= 53.97 μF
To determine the capacitance of capacitor B, we need to start by understanding the relationship between the energy stored in a capacitor and its capacitance.
The energy stored in a capacitor can be calculated using the formula:
E = (1/2) * C * V^2
Where:
E is the energy stored in the capacitor,
C is the capacitance of the capacitor,
V is the voltage across the capacitor's plates.
Given that capacitor A and capacitor B have the same voltage across their plates, we can equate their energies:
E A = E B
Using the energy formula for capacitor A:
E A = (1/2) * C A * V^2
We can now substitute the values into the equation and solve for C B:
E B = (1/2) * C B * V^2
Since the energy of capacitor A can melt m kilograms of ice at 0 °C and the energy of capacitor B can boil the same amount of water at 100 °C, we can equate the energies:
Lf * m = Lv * m
Substituting the energy equations for capacitor A and B, we have:
(1/2) * C A * V^2 = (1/2) * C B * V^2
Canceling out the common terms and rearranging the equation, we get:
C B = C A
Therefore, the capacitance of capacitor B is equal to the capacitance of capacitor A, which is 8.4 μF.