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"A 4.74 kW kettle holds 1kg of water at a room temperature of 21 °C. How long would it take the kettle to bring all the water to a boiling point of 100 °C? Assume the kettle is 100% efficient and the specific heat capacity of water is 4200 j/kg."

In this scenario, we’re essentially witnessing a change in the internal energy of the water. More specifically it is an increase in kinetic energy we’re talking about, since we’re seeing an increase in temperature, brought about by the addition of heat into the system.

Let us first work out the amount of heat energy added, using the formula:
Q = mcΔT
Where c = specific heat capacity
Q = heat energy
m = mass
ΔT=change in temperature
Since m = 1 kg, c = 4200 J/kg°C, and T = 100 - 21 = 79°C therefore:
Q = 1 . 4200 . 79 = 331800 J
Now our next step is to find the time taken to supply this amount of energy to the water. Since we’re assuming 100 % efficiency, we can assume there’s no extraneous energy change like sound or light. In other words:
Electrical energy input from kettle = Heat energy output to water
Therefore, the amount of electrical energy supplied to the kettle is also 331800 J. Now all that is left is to find the power supply to the kettle with the formula:
P = E/t  U+21D2.svg  t = E/P
Since E = 331800 J and P = 4.74 x 103 W, therefore:
t = 331800/4.74  103 = 70 s

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