Question

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.

Answer 1

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

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

Answer 2

Heya....

===== Answer =====

Power of heat 20 kW

heat of water = 4.186 KJ/kg

mass of water = 20 kg

∆ T = 10 to 30°C

time taken = t


Heat is taken by water is ---

P X T = mc∆T

2 X t = 20 X 4.186 X ( 30-10)

2t = 1674.4

t = 1674.4/2

= 8372.2 sec...

Thank you