All the heat generated by a current of 2A passing through a resistor of 6Ω for 2.5s is used to evaporate 5 g of a liquid at its boiling point. What is the s...
Question 1 Report
All the heat generated by a current of 2A passing through a resistor of 6Ω for 2.5s is used to evaporate 5 g of a liquid at its boiling point. What is the specific latent heat of the liquid?
Answer Details
The heat generated by the current passing through the resistor is given by the formula H = I2RT, where H is the heat energy, I is the current, R is the resistance, and T is the time for which the current flows. In this case, I = 2A, R = 6Ω, and T = 2.5s. So, the heat generated is: H = (2A)2 x 6Ω x 2.5s = 60J This heat is used to evaporate 5g of a liquid at its boiling point. The specific latent heat of vaporization (L) of the liquid is given by the formula: L = H/m where m is the mass of the liquid. In this case, m = 5g, so: L = 60J / 5g = 12J/g Therefore, the specific latent heat of the liquid is 12J/g. The option closest to this value is 120 Jg-1, so that would be the answer.
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