Answer to Question #264970 in Molecular Physics | Thermodynamics for Moonlight

Question

An immersion electric heater was put in some water and raises it's temperature from 40°c to 100°c in 6 minutes. After another 25 minutes it is noticed that half of the water has boiled away. Neglecting heat losses to the surrounding, calculate the specific latent heat of vaporisation of water.

Answer 1

Solution;

Let the rate of heat transfer from the heater to the water and QJ/min and total mass of water be 2m kg.

Specific heat capacity of water is $4200Jkg^{-1}K^{-1}$

The heat energy required to raise toe temperature of 2m kg of water from 40°c to 100°c is given as;

$=2m×4200(100-40)$

Now the amount of heat given by the heater in 6 minutes is;

$Q×6=2m×4200×60 J$ ....(1)

Again the amount of heat taken by water at 100°C for its vaporisation of half of its mass is (m×L)J,where L represents the latent heat of vaporisation.

This amount of heat is supplied by the heater in 25 minutes;

$Q×25=m×L$ ....(2)

Divide (2) by (1);

$\frac{m×L}{2m×4200×60}=\frac{25×Q}{6×Q}$

Hence;

$L=2100kJkg^{-1}$