It is valid to approximate temperatures to get a good general idea and a fairly accurate account of how efficient the kettle is and to outline whether the kettle is economically friendly or whether a new kettle should be purchased. On the other hand to be able to place an exact efficiency rating to help sell the product, I believe it is much more valid if the actual temperature was measured precisely for a more exact and accurate figure. I don’t feel the results would be that much dissimilar but there would be little room for questioning accurate measurements.
- Is it valid to assume that the specific heat capacity for water is constant from 15 °C up to 100 °C?
Yes, because the specific heat capacity for water is constant between 15˚C up to 100˚C in liquid form, the specific heat changes when the liquid becomes a gas/steam.
- Can you think of any other experimental errors that may have affected the results and how much they are likely to affect the calculations?
This is not really relevant in the experiment I did, but if the water was boiled at a different altitude the time taken to reach boiling point would have differed, and this could differ by a lot. In my experiment the temperature of the air could affect the time taken because the water will be cooler at room temperature than it would be in the summer, this could cause small variation in time taken to bring the water to boil. Furthermore it is not sure when the kettle was last boiled, there could have been a little heat energy left within the kettle, this could change the result slightly if the kettle had been boiled fairly recently.
Experiment 2 – Results and calculations
Calculating the efficiency of your microwave
- From the experiment you have performed, explain in no more than 150 words the purpose of the investigation.
The investigation is designed to investigate and determine the efficiency of a microwave at bringing water to the boil, which will help to determine the amount of energy that is wasted when the microwave boils the water to its boiling point, showing how effective the product is at doing the required task. This allows for comparison between boiling water in the microwave compared with the kettle.
- What was the rating of your microwave? Include the correct units.
The rating of the microwave was 900W (Watts).
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Explain how you measured the mass you used for the calculation and the value of mw you came up with. Include the correct units.
I used the same mass of water, which I wanted to bring to boil as the kettle experiment 300g (Grams) this was measured the same way using a measuring jug and converting millilitres to grams.
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What energy Qw was calculated to bring your mass of water to the boil? Include the correct units.
It took 106845J (joules) of energy required to bring 300ml of water to boil.
- How long did it take to bring the water to the boil? Include the correct units.
It took a154s (seconds) for the microwave to boil 300ml (Millilitres).
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What energy input Qmv was required from your microwave? Include the correct units.
The energy input from the microwave was 138600J (Joules).
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What efficiency ηmv did you calculate? Does it compare favourably with other similar devices (you may want to search online for the efficiency of similar microwaves)?
The efficiency of the microwave was 77.1% which is higher than the average 64% for a microwave, although there are some microwaves with efficiency greater than 70%. So the microwave has a good efficiency level compared with other products and the efficiency is greater than that of the kettle.
- Note any observations you made in performing the experiment.
The experiment was difficult to decide whether the water had boiled as much as it had in the kettle, also did not want to spill water all in the microwave which could have affected the results. The water seemed to take a lot longer to boil in the microwave.
- Account for any energy losses that have occurred.
There are losses that are created by the microwave having a light inside and the power needed to rotate the moving plate which will use energy even though they are not used for the actual heating. Heat will be lost to the glass on the door which in turn will lose heat to the atmosphere; heat will be lost to the materials inside the microwave.
- Can you think of any other experimental errors that may have affected the results and how much they are likely to affect the calculations?
It was extremely difficult to tell whether the water had boiled, the same amount as in the kettle, which could affect the calculations because if the water had not boiled to the same, the time taken would be less than the actual amount required giving a different answer to what it should be. Same as the kettle the room temperature will affect the result because if the temperature is cooler there will be more heat transfer from the microwave to the atmosphere and less if the temperature is more.
- If you look through the calculation spreadsheet, you will see that assumed values of efficiency are used for an electric kettle and a gas hob or electric ring. The kettle has an efficiency of approximately 85%, which is a typical value. A gas hob has an efficiency of approximately 35%, which is the same as an electric ring. A microwave has an efficiency of approximately 60%. Based on these values, write down your preferred method for boiling water, putting them in order of energy consumption with the most energy efficient first and the worst one last.
From the percentages outlined in the question my preferred method of boiling the water would be to use the kettle because as it has 80% efficiency, then the microwave because that has a 60% efficiency and last I would use the gas hob as that as the least efficiency with only 35%. This is based on the efficiency of the product alone.
- For each of the three methods above, explain why the efficiencies vary so much and list the types of losses involved with each method.
The efficiency varies a lot because the methods are different, the losses associated with each method differ, and some methods lose more energy than others because the stages of transfer differ.
The kettle has its heating element inside the water so any heat that wants to escape must transfer through the water before being able to transfer else were, although there are loses to the casing of the kettle and then to atmosphere but it must travel through the water to reach the casing then the atmosphere. There are losses in energy as the kettle makes noise. Also a lot of heat energy is allowed to escape as steam through the top.
The microwave is not quite as efficient as the kettle as the heating element is not within the water the waves must travel to the water, although the heat is sealed as well as possible within the space inside the microwave. There are energy loses to the cup that the water is held in, there are losses to the material of the microwave and also loses to the atmosphere through the glass on the microwave door.
The gas hob is the least efficient, this is because the heat energy can escape much easier to the atmosphere as there is nothing to prevent this, and furthermore the heat energy from the burning gas must also pass upwards through the base of the pan to reach the water which uses lots of energy. So there are lots of loses to the environment, to the pan containing the water and energy is also lost as steam.
- Do some research to find typical values for electricity and gas costs per unit. Units are in kWh. Based on these findings and the efficiencies given above, which method of water heating do you think is the least expensive (gas hob, electric kettle, or microwave) and give reasons why with calculations to back up your choice?
e-on.
Electric: first 900kWh (kilowatt an hour) 24.8325p (Pence)
Then 12.5475p per kWh (Pence per kilowatt an hour)
Gas: first 2680kWh (kilowatt an hour) 7.7679p (Pence)
Then 3.37995p per kWh (Pence per kilowatt an hour)
The cost to boil the kettle is (2 x 0.02389 = 0.04778 kWh) which costs 0.76p.
The cost of using the microwave is (0.9 x 0.04278 = 0.0385kWh) which costs 0.48p.
An estimated cost for the gas hob is (1.0 x 0.0833 = 0.0833 kWh) which costs 0.28p.
The above calculations show that even though the kettle and microwave are far more efficient than that of the gas hob, it still works out cheaper and is more cost effective to use the less efficient item to boil the required 300ml (Millilitres) of water. So because of the price of gas compared with that of electricity it is far cheaper to use the gas hob. Which does raise the question whether efficiency is the cheapest method in this case the answer is no.
Experiments 3 and 4 – Results and calculations
- What mass of ice cube did you use? Include the correct units.
The mass of the ice cubes I used were 12.5g (Grams) and 15g (Grams).
- How long did it take for the ice cube to melt in the mug of boiled water?
The time taken to melt the ice cube was 39s (Seconds) in the mug of boiling water and this was for the 12.5g (Grams) ice cube to completely melt.
- What heat transfer rate did you calculate for this process? Include the correct units.
107W (Watts) was calculated for the transfer rate for the ice cube to melt in a mug of boiling water.
- How long did it take for the ice cube to melt in the microwave?
The time taken to melt the ice cube in the 900W (Watts) microwave was 118s (Seconds) and this was for the 15g (Grams) ice cube to fully melt.
- Compare this with the time taken in experiment 3 and comment on the results.
The ice cube took longer to fully melt in the microwave than it did in the mug of boiling water, I had expected this as the microwave will take time to warm up and the ice cube was bigger than the one in the cup, although it did take longer than I imagined it would take to melt the ice cube.
- Compare the heat transfer rate of the microwave when melting ice with that found for the mug of boiled water.
The heat transfer rate was a lot quicker between the mug of boiling water to melt the ice cube than that of the microwave to melt the ice cube.
- What efficiency did you calculate for the melting process using the microwave?
4.7% efficiency was calculated for the melting of the ice cube in the microwave.
- Compare this with the figure found in experiment 2 and comment on the results.
77.1% efficiency was calculated in experiment 2 for the boiling of the water whereas only a 4.7% for the melting of ice which is a drop of 72.4% efficiency. This is a very large drop in efficiency suggesting a lot more energy is required from a microwave to melt ice than to boil water.
- From the final few answers, you have hopefully found that something strange happens when ice is melted in a microwave. In theory, the higher power of the microwave should have melted the ice cube a lot more quickly than the mug of boiled water. The efficiency of the microwave for melting probably also came out a lot lower than that for boiling. Carry out some research of your own and see if you can discover the reasons why the way in which a microwave operates affects the efficiency of the heating process and depends on the phase of the water present. Present your findings here in no more than 200 words.
From the research I carried out, I found that the waves of the microwave react to water vapour which in turn makes it easier to warm water than it does ice, ice must go through an extra stage of turning the ice into water, which will then allow the ice to melt quicker, which was visible in the experiment because it seemed to take a while to produce a little bit of water but when water was present the melting speeded up. So water absorbs the microwaves easier. Furthermore the hydrogen bonds of ice are stronger and more compact making in more difficult for the microwaves to break these bonds and cause vibration of the particles, this vibrating energy causes heat.
