Hi it's Mr Rees from Mansbury School Science and I'm back with another GCSE Required Practical to show you how to measure the specific heat capacity of a material. Now I have three materials here today. I have brass, I have copper and aluminium and I've already got my aluminium set up to go.
What I have is a heater that just pops into the hole in the block. There's also a smaller hole for the thermometer to go. Now you need to make sure that you have a couple of drops of water in there first because otherwise there's going to be an air gap between the metal and the thermometer and you won't get an accurate reading.
We're going to turn the heater on and record the temperature every minute for 10 minutes. Now I have my power supply here. We're going to use DC and we're going to use about 10 volts. It doesn't need to be exactly 10 volts, but that's a good number to aim for. Now I'm going to take a reading of the temperature before I start and getting on eye level.
I can see the temperature is 24 degrees Celsius. I'm going to start my stop clock as soon as I turn my power supply on. So the equation for specific heat capacity is energy, or heat transferred measured in joules, is equal to mass in kilograms, or it could be grams, times SHC for short, that's specific heat capacity, times temperature change in degrees Celsius. Now rearranging the equation, we have SHC equals... Energy divided by mass times temperature change.
So therefore the unit of SHC is joules per kilogram or gram per degree Celsius. So that tells us what SHC is. It's the energy needed in joules to raise the temperature of one kilogram of a material by one degree Celsius. It is going to warm up slowly to begin with because first of all the heater has to heat up before it can heat up the block.
at a steady rate. So in order to calculate specific heat capacity we need to know the mass of our block and these have been specially made to be one kilogram. We need our temperature change and that's what we're recording every minute but we also need to know the energy and the way we're going to calculate energy is using power. So we need to know the power that is going into our heater.
Now there's two different ways of doing this. There's the really easy way and that's using a dual meter. What you can do is count how many flashes you have in 30 seconds, and in my case it's flashing 19 times in 30 seconds, and each flash is 100 joules, so 19 times 100 is 1900 joules in 30 seconds.
So to calculate power, that's energy per second, joules per second, we take 1900, divide that by 30, and that gives us 63 joules per second. The proper unit for power is watts, so that's 63. watts. Now this is the really easy way of calculating power but it's not very accurate because this is measuring the power going into the power supply. There might be some power lost in the power supply so not all of that is going to go into the heater.
So there's a more accurate way of measuring power and that's with a voltmeter and an ammeter. So the more accurate way of measuring power is by multiplying the current flowing through the heater by the PD or potential difference or voltage across it. So first of all we're going to pop in our ammeter to measure the current in series with the heater. So ammeters always go in series with the component that we're looking at. Then we have our voltmeter and voltmeters always go in parallel across the component that we're looking at.
So we take our two leads from our voltmeter and they always piggyback onto the leads going into the components. So now we have a voltage or a PD, potential difference across our heater, and the current flowing through it. Now we're not really concerned with the voltage shown on the power pack because that's not going to be accurate. We need to be looking at the voltmeter.
You can change the output from the power pack to make sure that you get a number nearer 10 volts if you want. Now your PD and your current might fluctuate throughout the experiment and it's up to you whether you want to record your voltage and your current every minute as well and then you can take a mean of all of those. but they're not going to change a huge amount.
So I would just take the 10 volts that we have on here, and I can see that the current is 3.47 amps. So power equals current times voltage. So that's 10 times 3.47. That gives us 34.7 watts.
And as you can see, that's a much lower number than the 63 that we had earlier, but this is a much more accurate value. So here are my results after 10. minutes. As you can see the block started heating up a lot more quickly.
Now we need to do a graph of our results. What we do is put temperature on the y-axis and we put time on the x-axis. Now you can if you want to put energy on the x-axis instead, but that just means that you have to take your power and times it by time for every single one of your results.
So the easier thing to do is just put time on the x-axis. Now if we rearrange the equation to get specific heat capacity we have SHC equals power times time divided by mass times temperature change. Now what we can do is use our linear portion of our graph to get time divided by change in temperature. What we need to do is make a right angle triangle with our line of best fit for the linear part of our graph. And then we're going to take the width of the triangle, that's our change in time, and divide it by the height of the triangle, that's our change in temperature.
If we take this number and then we times it by the power, which for me was 34.7 and then divide by the mass which in my case was one kilogram that gives us the specific heat capacity. For me that ends up being 1050. Now the actual SHC for aluminium is 900 joules per kilogram per degree Celsius. That means that if you have one kilogram like this it needs 900 joules to raise the temperature by one degree Celsius.
Now our number is quite a bit bigger than that. Now why is that? Well it's because this is not insulated so as energy is going into the aluminium block energy is also being lost to its surroundings. So it appears to us that we need more energy than we should in reality. One way around this is to insulate the block.
We could use foam on the sides and especially on the bottom. What you can then do is repeat the experiment for different materials like copper and brass and see how their specific heat capacities compare.