Question

All the Heat generated by a current of 2Ampere passing through a 6ohms resistor for 25seconds is used to evaporate 5g of liquid at its boiling point. What is the specific Latent Heat of the Liquid?

Answer 1

The energy balance equation

$$I^2Rt=Lm$$

So, the specific latent yeat of the liquid

$$L=\frac{I^2Rt}{m}$$

$$=\frac{(2\:\rm{A})^2\times 6\:\Omega\times 25\:\rm{s}}{0.005\:\rm{kg}}=120\:000\;\rm{J/kg}$$