An electric heater put in some water raises the temperature of the water from 40°C to 100°C in 6 minutes. After another 25 minutes, it's noticed that half the water has boiled away. Neglecting heat losses to the surrounding, calculate shc of vaporisation?

1 Answer
Feb 26, 2018

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

Now heat energy required to raise the temperature of #2m# kg water from #40^@C# to #100^@C# will be

#=2m*4200*(100-40)# J

#=2m*4200*60# J
, Where #4200Jkg^-1K^-1# is the shc of water.
Now the amount of heat given by the heater in 6min

#Q*6=2m*4200*60.....(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 #LJkg^-1# represents the latent heat of vaporisation.

This amount of heat is supplied by the heater in 25min

So #Q*25=m*L.......(2)#

Dividing (2) by (1) we get

#(mL)/(2m*4200*60)=(Q*25)/(Q*6)#

#=>L=25/6*2*4200*60#

#=>L=21*10^5Jkg^-1=2100kJkg^-1#