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?
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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.
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