Answer to Question #117945 in Physics for Arogundade Moses Temitope

Question

An electric kettle has a 750W-240V heater and is used on a 200V mains. if the heat capacity of the kettle is 400J/K and the initial water temperature is 20°C, how long will it take to boil 500g of water, assuming the resistance of the heater is Unaltered on changing to the new mains.

Answer 1

First, obtain the power of the kettle that is used at 200V-circuit:

$R=\frac{V_1^2}{P_1}=\frac{V^2_2}{P_2},\\ \space\\ P_2=P_1\bigg(\frac{V_2^2}{V_1^2}\bigg)^2.$

The amount of heat, on the one hand, is

$Q=P_2t,$

and, on the other hand, it is the heat required to heat the water and kettle up.

$c$ is specific (per 1 kg of substance) heat capacity of water, 4200 J/(kg K).

$m$ is mass of the water, 0.5 kg.

$C$ is heat capacity of the kettle, 400 J/K.

Indeed, to boil the water, we must heat 0.5 kg of water from 20 to 100°C. The heat required for this is

$Q_w=cm\Delta \tau.$

Also, heating the kettle from 20 to 100°C requires thermal energy

$Q_k=C\Delta \tau,$

therefore, the total energy required is

$Q=Q_w+Q_k=\Delta\tau(cm+C). \\ \space\\ t=\frac{V_1^2\Delta \tau(cm+C)}{V_2^2P_1}=384\text{ s}.\\$

It is equivalent to 6 minutes and 24 seconds.