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
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