How much energy does it take to boil a kettle?

As an expert in thermodynamics and energy efficiency, I can provide a detailed explanation on the energy required to boil a kettle. Boiling a kettle is a common household task that involves converting electrical energy into thermal energy to heat water. The amount of energy required to boil a kettle depends on several factors, including the volume of water, the initial temperature of the water, the desired final temperature, the efficiency of the kettle, and the type of heating element used.

To begin with, let's consider the average kettle capacity. A typical kettle holds around 1.5 pints, which is approximately 840 milliliters. The energy required to heat water is determined by the specific heat capacity of water, which is the amount of heat needed to raise the temperature of one unit of mass by one degree Celsius. For water, this value is approximately 4.186 joules per gram per degree Celsius (J/g°C).

The energy (\( Q \)) required to heat a certain mass (\( m \)) of water by a temperature change (\( \Delta T \)) can be calculated using the formula:
\[ Q = m \times c \times \Delta T \]
where:
- \( m \) is the mass of water in grams,
- \( c \) is the specific heat capacity of water (4.186 J/g°C),
- \( \Delta T \) is the change in temperature in degrees Celsius.

Assuming the initial temperature of the water is room temperature, which we can approximate as 20°C, and we want to bring the water to a boil at 100°C, the temperature change (\( \Delta T \)) would be 80°C.

If we take the average kettle capacity of 1.5 pints and convert it to grams (knowing that 1 pint is approximately 568 grams), we have:
\[ 1.5 \text{ pints} \times 568 \text{ grams/pint} = 852 \text{ grams} \]

Now, we can calculate the energy required to heat this amount of water:
\[ Q = 852 \text{ g} \times 4.186 \text{ J/g°C} \times 80°C \]
\[ Q = 271,372.8 \text{ joules} \]

This is the theoretical energy required if all the electrical energy were perfectly converted into heat without any losses. However, in reality, no heating system is 100% efficient. A typical electric kettle has an efficiency of around 80-90%. Let's assume an efficiency of 85%. This means that for every 100 joules of electrical energy used, only 85 joules are effectively used to heat the water.

To find the actual electrical energy required, we divide the calculated thermal energy by the efficiency:
\[ \text{Electrical Energy} = \frac{271,372.8 \text{ joules}}{0.85} \]
\[ \text{Electrical Energy} = 320,434.5 \text{ joules} \]

This is the energy in joules. To convert it to kilowatt-hours (kWh), which is a more common unit for measuring electrical energy, we use the conversion factor:
\[ 1 \text{ kWh} = 3,600,000 \text{ joules} \]

So, the electrical energy required to boil one full kettle of water is:
\[ \text{Electrical Energy} = \frac{320,434.5 \text{ joules}}{3,600,000 \text{ joules/kWh}} \]
\[ \text{Electrical Energy} \approx 0.0891 \text{ kWh} \]

Now, regarding the cost, if electricity costs 15p per kWh (as an example), the cost to boil one full kettle would be:
\[ \text{Cost} = 0.0891 \text{ kWh} \times 15 \text{ p/kWh} \]
\[ \text{Cost} \approx 1.34 \text{ p} \]

This is a more detailed and accurate calculation than the provided reference, which seems to have some inaccuracies in its assumptions and calculations.

The average kettle holds 1.5 pints and uses about one unit of electricity to boil 12 pints of water (or 8 x 1.5 pint-full kettles) -C so that's around 2.5p every time you boil a full kettle. The average slow cooker uses about one unit of electricity or 20p to power eight hours of cooking -C that's roughly 2.5p per hour.Feb 1, 2013