Could anyone help me with part b of this physics question? I'm just confused what equation I should use for part b to find energy transfer. Any help ASAP would be very much appreciated!

Question

The photograph shows a 'quiet boil' electric kettle. The makers of the kettle claim that it boils water with much less noise than a standard kettle.

 

A laboratory technician takes some measurements to compare a 'quiet boil' electric kettle with a standard electric kettle.

The table shows the results recorded by the technician.

Quiet boil kettle data

Mass of water = 1.2kg

Initial temperature of water = 10degreesC

Final temperature of water = 100degreesC

Potential Difference= 243V

Current = 11.9A

Time taken to heat water to boiling point - 168s

Average sound Intensity = 3.72mWm^-2

Standard Kettle Data

Mass of water = 1.2kg

Initial temperature of water = 10degreesC

Final temperature of water = 100degreesC

Potential Difference= 247V

Current = 11.8A

Time taken to heat water to boiling point - 172s

Average sound Intensity = 10.5mWm^-2

(a) A student uses the values in the table to calculate the efficiency of each kettle at heating the water to boiling point. He calculates the efficiency of the 'quiet boil' kettle to be 0.93

Calculate the efficiency of the standard kettle.

specific heat capacity of water = 4180 J kg⁻¹ K⁻¹ (4 marks)

(b) The intensity of the sound produced by each kettle was measured with a sound meter which was 30.0 cm from the centre of the kettle.

Calculate the energy transferred by sound while the water in the standard kettle is brought to the boil. You may treat the kettle as a point source. (4 marks)

(c) The label on the original packaging of the quiet boil electric kettle states, 'This kettle is much more efficient than a standard kettle because it produces less sound.' (2 marks)

Explain the extent to which this statement is supported by your calculations.

 

Answer 1

The average Power of the source = 4πr²I where r is the radius of the sphere around the source. In this case the distance to the sound meter in meters. The units of the power will be in mWatts if you use the intensity given.

The sonic energy is just the average power * time in seconds (in mJoules unless you convert to Watts beforehand)