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A water heater marked 2 kW, 220 V is connected to a 220 V supply line. How long will it take to heat 20 kg of water from 10°C to 30°C? Assume that all the

heat is taken up by the water.



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Power of heater(P) = 2kW
Specific heat of water (c)= 4.186 kJ/kg
mass of water (m)= 20 kg
temperature change (ΔT)= 10°C to 30°C
let the time taken = t second

As all energy is taken up by the water, we just need to equate the amount of heat released by water in t seconds to the amount of heat required to raise the temperature from 10
°C to 30°C. Thus

P*t=mc \Delta T\\ \\2*t=20*4.186*(30-10)\\ \\2t=1674.4\\ \\t= \frac{1674.4}{2}=\boxed {837.2\ seconds}
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