A 400W immersion heater is placed in a pot containing 2 litres of water at20C.(a)How long will it take to bring the water to boiling temperature, assuming that 80% of the available energy is absorbed by the water?(b)How much longer will it take to boil half the water away?
Let's assume that the time taken to heat water upto boliing temperature (100 C) is `t_1` .
The specific heat capacity of water can be assumed as constant through out the whole temperature range and it is 4.187 kJ/kg.K and latent heat of evaporation is 2270 kJ/kg. Density of water also assumed to be 1000 kg/m^3 throughout the temperature range.
Then if only 80% of heat from heater is absorbed.
`0.8 xx400 W xx t_1 = 2 xx 10^(-3) m^3 xx 1000 kgm^(-3) xx 4187 J/(kgK) xx (100 -20) K`
`t_1 = 2093.5 ` seconds.
The time taken to heat water upto boling temperature is 2093.5 seconds (34.89 minutes).
If the time taken to boil away half of water from `t_1` is `t_2` , then,
`0.8 xx 400 W xx t_2 = 1/2 xx 2 xx 10^(-3) m^3 xx 1000 kgm^(-3) xx 2270 xx 10^3 J/(kg)`
`t_2 = 7093.75` seconds
Therefore total time taken to boil away half of water is, T,
`T = t_1 + t_2 = 2093.5 + 7093.75 = 9187.25` seconds.
or 2 hours and 33 minutes.