Question on: JAMB Physics - 2024
5 x 10-3 kg of liquid at its boiling point is evaporated in 20 s by the heat generated by a resistor of 2 Ω when a current of 10 A is used. The specific latent heat of vaporization of the liquid is
A
8.0 x 104 J/kg
B
8.0 x 105 J/kg
C
8.0 x 106 J/kg
D
8.0 x 107 J/kg
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Correct Option: B
Here's how to solve the problem:
- Calculate the heat energy generated by the resistor:
- The power (P) dissipated by a resistor is given by P = I²R, where I is the current and R is the resistance.
- P = (10 A)² * 2 Ω = 200 W
- The heat energy (Q) generated in 20 seconds is Q = P * t, where t is time.
- Q = 200 W * 20 s = 4000 J
- Calculate the specific latent heat of vaporization (L):
- The heat energy (Q) used to evaporate the liquid is given by Q = mL, where m is the mass of the liquid and L is the specific latent heat of vaporization.
- L = Q / m
- L = 4000 J / (5 x 10-3 kg) = 8.0 x 105 J/kg
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