WEBVTT
00:00:00.320 --> 00:00:07.520
An iron cylinder has a mass of 0.80 kilograms and a temperature of 1.00 times 10 to the power of three degrees Celsius.
00:00:08.010 --> 00:00:13.470
The cylinder is dropped into an insulated chest containing 1.0 kilograms of ice at its melting point.
00:00:14.420 --> 00:00:17.530
What is the equilibrium temperature of the chestโs contents?
00:00:18.050 --> 00:00:34.490
Use the value of 4180 joules per kilogram degrees Celsius for the specific heat capacity of water, a value of 334 kilojoules per kilogram for the latent heat of fusion of water, and a value of 452 joules per kilogram degrees Celsius for the specific heat capacity of iron.
00:00:35.280 --> 00:00:43.510
Okay so in this question, there are three objects that we need to think about: firstly, the iron cylinder; secondly, the ice; and thirdly, the chest which contains the ice.
00:00:43.950 --> 00:00:45.690
Letโs start by thinking about the iron cylinder.
00:00:46.160 --> 00:00:47.680
So here is our iron cylinder.
00:00:48.040 --> 00:00:51.470
Now letโs write down all the information that we have about this cylinder from the question.
00:00:51.940 --> 00:00:58.010
Firstly, we can say that it has a mass, which weโll call ๐ sub ๐ for mass sub iron, of 0.80 kilograms.
00:00:58.640 --> 00:01:03.840
Secondly, it has a temperature ๐ sub ๐ of 1.00 times 10 to the power of three degrees Celsius.
00:01:04.490 --> 00:01:06.740
In other words, this is 1000 degrees Celsius.
00:01:07.190 --> 00:01:14.190
And finally, we know that it has a specific heat capacity which, weโll call ๐ sub ๐ of 452 joules per kilogram degrees Celsius.
00:01:14.640 --> 00:01:14.920
All right!
00:01:14.920 --> 00:01:16.100
So thatโs the iron cylinder.
00:01:16.420 --> 00:01:19.490
Letโs now think about the ice thatโs in the chest okay.
00:01:19.490 --> 00:01:21.030
So here are our cubes of ice.
00:01:21.290 --> 00:01:26.000
Now we know that the ice thatโs in the chest has firstly a mass of 1.0 kilograms.
00:01:26.440 --> 00:01:31.310
And by the way weโre calling the mass ๐ sub ๐ค because weโve already used the subscript ๐ to represent iron.
00:01:31.690 --> 00:01:34.890
So we use the subscript ๐ค to represent water rather than ice.
00:01:35.470 --> 00:01:39.170
Anyway, so weโve also been told in the question that the ice is at its melting point.
00:01:40.000 --> 00:01:45.730
So the temperature of the ice, initially, ๐ sub ๐ค is zero degrees Celsius, because thatโs the melting point of ice.
00:01:46.150 --> 00:01:51.780
Thirdly, weโve been told that the specific heat capacity of water is 4180 joules per kilogram degrees Celsius.
00:01:52.240 --> 00:01:58.640
And finally, weโve been told that the latent heat of fusion of water, which weโll call ๐ฟ sub ๐ค, is 334 kilojoules per kilogram.
00:01:59.130 --> 00:02:01.870
So thatโs all the information weโve got about the ice from the question.
00:02:02.350 --> 00:02:04.740
Now letโs finally think about the insulated chest.
00:02:05.210 --> 00:02:09.070
Now we donโt have a lot of information about this chest, but we do know that itโs insulated.
00:02:09.520 --> 00:02:16.210
This means that once we drop the iron cylinder into the chest, the iron cylinder can only interact with the ice in the chest.
00:02:16.700 --> 00:02:23.300
Because the chest is insulated, the only heat transfer is between the contents of the chest, in other words, the iron cylinder and the ice.
00:02:23.650 --> 00:02:27.020
The two can interact with each other, but not anything outside of the chest.
00:02:27.440 --> 00:02:30.730
And so there is no heat transfer either into the chest or out of the chest.
00:02:31.250 --> 00:02:37.490
And with all of this in mind, we need to try and find the equilibrium temperature once the iron has been dropped into the chest of ice.
00:02:38.150 --> 00:02:40.840
Okay, so at this point weโve got all the information we can from the question.
00:02:40.840 --> 00:02:41.970
So letโs clear it off the screen.
00:02:42.430 --> 00:02:48.420
So now letโs take our iron cylinder and drop it into our chest, and then letโs close our chest thus making it fully insulated.
00:02:48.780 --> 00:02:53.800
Now the thing that we need to remember is that objects in thermal equilibrium are at the same temperature.
00:02:54.240 --> 00:03:05.370
In other words, even though the iron starts out at a temperature of 1000 degrees Celsius and the ice starts out at a temperature of zero degrees Celsius, theyโre both gonna end up at the same temperature once theyโre in thermal equilibrium.
00:03:05.730 --> 00:03:16.760
And so weโll say that the temperature of the iron at the end of the process, which weโll call ๐ sub ๐ superscript fin, is going to be equal to the temperature of the water at the end of the process as well.
00:03:17.600 --> 00:03:21.100
So thatโs one important piece of information that weโll need in order to solve our problem.
00:03:21.630 --> 00:03:27.370
Now we can relate these final temperatures of the iron and of the water to the initial temperatures of the iron and water.
00:03:27.870 --> 00:03:40.610
Letโs say that the final temperature of the iron is equal to the initial temperature of the iron, which we know to be 1000 degrees Celsius, plus the change in temperature that the iron undergoes in this process, which weโll call ฮ๐ sub ๐.
00:03:41.110 --> 00:03:50.770
Similarly, weโll say that the final temperature of the water is equal to the initial temperature of the water, which happens to be zero degrees Celsius, plus the change in temperature that the water undergoes.
00:03:51.260 --> 00:03:57.090
But then this very first equation is telling us that the final temperature of the iron is the same as the final temperature of the water.
00:03:57.550 --> 00:04:02.870
Therefore, this is equal to this, and so we can equate the two right-hand sides of both of these equations.
00:04:03.360 --> 00:04:13.490
So instead of talking in terms of ๐ sub ๐ superscript fin and ๐ sub ๐ค superscript fin, letโs just say that ๐ sub ๐ plus ฮ๐ sub ๐ is equal to ๐ sub ๐ค plus ฮ๐ sub ๐ค.
00:04:13.920 --> 00:04:28.700
Now we can intuitively realize that one of these ฮ๐ sub ๐ or ฮ๐ sub ๐ค is going to be negative, because whatโs going to happen is that heat is going to be transferred between the iron and the ice such that one of them increases in temperature and the other one decreases in temperature.
00:04:29.210 --> 00:04:31.130
But weโll come to that when we actually do the maths.
00:04:31.420 --> 00:04:33.010
For now, letโs consider something else.
00:04:33.380 --> 00:04:41.040
Letโs realize that because the chest is insulated, there is going to be no heat transfer between the contents of the chest and the external environment of the chest.
00:04:41.430 --> 00:04:45.950
In other words then, we can say that any change in the heat of the contents of the chest is equal to zero.
00:04:46.470 --> 00:04:54.580
But then, we can split up the heat transfer of the components of the chest into two parts: one is the heat transferred by the iron and the second is the heat transferred by the water.
00:04:55.020 --> 00:04:57.400
And when we add them together this is equal to zero.
00:04:57.960 --> 00:05:03.330
Now at this point, letโs look individually at the heat transfer of the iron and the heat transfer of the water.
00:05:03.770 --> 00:05:25.090
But before we do, letโs very quickly recall that the specific heat capacity of a substance is defined as the amount of heat change required per unit mass to increase the substanceโs temperature by one unit โ in this case, one degree Celsius โ and the latent heat of fusion of a substance is defined as the heat required to convert one unit mass from solid to liquid.
00:05:25.570 --> 00:05:28.900
Okay armed with that information then, letโs look at the heat transfer of the iron.
00:05:29.270 --> 00:05:32.390
Now in this situation, iron is not anywhere close to its melting point.
00:05:32.700 --> 00:05:34.490
So we donโt need to consider the melting of iron.
00:05:34.520 --> 00:05:37.110
We only need to consider the change in temperature of the iron.
00:05:37.580 --> 00:05:49.220
And having been given the specific heat capacity of the iron, we can then rearrange this equation for the iron to say that the heat transferred by the iron is equal to its mass multiplied by its specific heat capacity multiplied by its change in temperature.
00:05:49.790 --> 00:05:53.640
And so we can say that ฮ๐๐ becomes ๐๐๐ถ๐ ฮ๐๐.
00:05:53.980 --> 00:06:02.330
And then to this we need to add the heat transfer of the water, but this is a little bit more complicated, because remember weโve been told that the water is at its melting point.
00:06:02.700 --> 00:06:04.000
Thatโs zero degrees Celsius.
00:06:04.400 --> 00:06:09.880
Therefore, any heat transferred to the water, or ice in this case, will go towards melting the ice first.
00:06:10.350 --> 00:06:16.500
And then once all of the ice has completely melted, then any remaining heat will go towards increasing the temperature of the water.
00:06:17.000 --> 00:06:20.330
So we can take ฮ๐ sub ๐ค and split it into two parts.
00:06:20.660 --> 00:06:30.310
We can split it into the heat transferred into the ice that causes the melting of the ice plus the heat transferred to the water once all of the ice has melted that goes towards increasing the temperature of the water.
00:06:30.940 --> 00:06:48.930
Now this second bit, ฮ๐ sub ๐ค superscript temp, the heat transferred in order to increase the temperature of the water is going to have a similar expression as ฮ๐ sub ๐, because when weโre raising the temperature, the heat transferred is equal to the mass of the water multiplied by the specific heat capacity of the water multiplied by the temperature change of the water.
00:06:49.410 --> 00:06:55.140
However, to talk about the heat transferred to the ice that causes it to melt, we need to look at the latent heat of fusion.
00:06:55.490 --> 00:07:02.820
And once we rearrange this equation, we find that the heat transferred is equal to the latent heat of fusion of water multiplied by the mass of the water.
00:07:03.340 --> 00:07:06.890
And so we have an expression for ฮ๐ sub ๐ค superscript melt as well.
00:07:07.490 --> 00:07:12.020
So, letโs now sub in our expressions for ฮ๐ sub melt and ฮ๐ sub temp.
00:07:12.560 --> 00:07:19.700
Now at this point, we can see that in this equation the only things that we donโt know the values of are ฮ๐ sub ๐ and ฮ๐ sub ๐ค.
00:07:20.130 --> 00:07:23.640
And so it looks like weโve got one equation with two variables which we canโt solve for.
00:07:23.990 --> 00:07:29.500
Except we have another equation that tells us the relationship between ฮ๐ sub ๐ค and ฮ๐ sub ๐.
00:07:29.900 --> 00:07:37.480
So letโs rearrange this equation in order to solve for either ฮ๐ sub ๐ or ฮ๐ sub ๐ค and then substitute that back into this equation.
00:07:37.920 --> 00:07:39.890
So letโs say we solve for ฮ๐ sub ๐.
00:07:40.250 --> 00:07:45.190
Well then, the right-hand side of that equation becomes ๐ sub ๐ค plus ฮ๐ sub ๐ค minus ๐ sub ๐.
00:07:45.760 --> 00:07:54.440
Then letโs take the right-hand side of this expression and sub it in to ฮ๐ sub ๐ over here, at which point the only variable we now donโt know when this equation is ฮ๐ sub ๐ค.
00:07:55.110 --> 00:07:57.590
So letโs rearrange to solve for ฮ๐ sub ๐ค.
00:07:58.060 --> 00:08:12.250
When we do we should see that ฮ๐ sub ๐ค is equal to here we go: ๐๐๐ถ๐ multiplied by ๐๐ minus ๐๐ค minus ๐ฟ๐ค ๐๐ค all divided by ๐๐ค ๐ถ๐ค plus ๐๐๐ถ๐.
00:08:12.870 --> 00:08:16.530
Now at this point, we can substitute in all the quantities on the right-hand side of this equation.
00:08:17.080 --> 00:08:26.010
When we do, it looks something like this and, remember, we need to convert the latent heat of vaporization of water from kilojoules per kilogram into joules per kilogram.
00:08:26.350 --> 00:08:29.160
We do this by multiplying by a factor of 10 to the power of three.
00:08:29.600 --> 00:08:34.400
And then at this point, we can find a value for ฮ๐ sub ๐ค by evaluating the right-hand side of this equation.
00:08:34.900 --> 00:08:40.590
When we do, we find that ฮ๐ sub ๐ค is equal to 6.077 dot dot dot degrees Celsius.
00:08:41.070 --> 00:08:50.980
But because the lowest number of significant figures that we were given a quantity to in the question is two significant figures, thatโs ๐ sub ๐ค, then we need to round our answer to two significant figures as well.
00:08:51.370 --> 00:08:55.530
Now letโs be careful here, because the temperature of the water weโve said is zero degrees Celsius.
00:08:55.590 --> 00:08:57.060
Isnโt that one significant figure?
00:08:57.350 --> 00:09:01.250
Well yes it is, but in the question all we were told was that the ice was at its melting point.
00:09:01.710 --> 00:09:06.300
So we could have very easily written this as 0.00000 degrees Celsius.
00:09:06.660 --> 00:09:10.910
And so the quantity with the lowest number of significant figures in the question was ๐ sub ๐ค.
00:09:11.510 --> 00:09:15.780
So anyway, ฮ๐ sub ๐ค becomes 6.1 degrees Celsius to two significant figures.
00:09:16.280 --> 00:09:18.010
Now weโve just had to do a lot of maths here.
00:09:18.010 --> 00:09:22.030
So letโs take a step back and think about what ฮ๐ sub ๐ค actually means.
00:09:22.370 --> 00:09:32.370
Well ฮ๐ sub ๐ค is the change in temperature of the ice or water once all of the heat exchange is complete between the iron and the ice and the two are in thermal equilibrium.
00:09:32.840 --> 00:09:39.950
Now because this value is positive, this means that the temperature of the ice by the end of the process will have increased by 6.1 degrees Celsius.
00:09:40.420 --> 00:09:46.470
Remember we said earlier that the final temperature of the water was equal to the initial temperature of the water plus the change in temperature of the water.
00:09:46.860 --> 00:09:51.520
Well, the initial temperature of the water was zero degrees Celsius, and we just worked out ฮ๐ sub ๐ค.
00:09:52.190 --> 00:09:59.980
Therefore zero degrees Celsius plus 6.1 degrees Celsius give us a value of 6.1 degrees Celsius for the final temperature of the water.
00:10:00.590 --> 00:10:03.890
Now we very easily couldโve worked out the value of ฮ๐ sub ๐ as well.
00:10:04.170 --> 00:10:06.140
And if we did that, weโd find it to be negative.
00:10:06.480 --> 00:10:12.200
Moreover, weโd also find that the final temperature of the iron cylinder was also 6.1 degrees Celsius.
00:10:12.570 --> 00:10:16.890
But then this shouldnโt surprise us, because this is what we stated was a condition of thermal equilibrium.
00:10:17.720 --> 00:10:23.230
But anyway, at this point, we have our final answer: the equilibrium temperature is 6.1 degrees Celsius.