Answer to Question #3265 in Molecular Physics | Thermodynamics for o m p

Question

A 2 kW kettle, 90% energy efficient is used to boil 1 liter of water from 20ºC to
100ºC. (heat capacity of water is 4180 J/kg.K, latent heat of water is 2256
kJ/kg)
i) How long does this take?(20marks)
ii) How long more to completely boil-off the water(5marks)

Answer 1

How long does this take?
P = A/t = Q/t → t_1 = Q/P = Q/np = (cm(100º-20º))/np = (4180*1(100º-20º))/(1.8*10³ ) = 334400/1800 = 185 sec.
How long more to completely boil-off the water.
t₂ = Q/P = Q/np = (cm(100º-20º)+qm)/np = (4180*1(100º-20º)+2256*1)/(1.8*10³ ) = 336656/1800 = 187 sec.