Answer to Question #290216 in Molecular Physics | Thermodynamics for Oluwatoyin

Question

A heater marked 60°c evaporates 6x10¯³ kg of boiling water in 60 seconds and what is the specific latent heat of vaporization of water in Jkg¯¹

Answer 1

Instead of 60°C it should be 60watt

H=mL

m=mass and L= specific latent heat of vaporization.

Also, H= Pt

P= 60watt, t= 60secs

$H= 60×60=6×10^{-3}×L$

$L=6×10^5Jkg^{-1}$