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?
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.