Teacher's Page for Heat Capacity

Teacher's Page for Measuring HEAT CAPACITY

To begin, you have to introduce the subject to your students. Being a subject that is at the fringes of students' awareness, it must be approached by going from they know to what they do not know. One suggested method you might look at is called "Safe, Hot Water at Your Campsite." Now let's turn to how you might conduct the laboratory experience.

|(Δtemp_water x mass_water)/(Δtemp_x x mass_x)| = C_rel = the SPECIFIC heat capacity of the thing you added

The smart teacher is interested in several parameters with regard to lab work: the least work for the least expense for the least amount of time without negatively impacting the lesson to be learned. The best way to add the most 'leasts' in this experiment is to use the fewest number of items to be assessed for heat capacity. The unwise teacher will test all sorts of things like iron, copper, brass, aluminum, and forget things that really impact the data. So how does the smart teacher make decisions as to what to include?

Take another look at Einstein's equation: it is loaded with constants. So what are the variables? "N" and "v"! Smart teachers recognize that denser things such as lead have lower values of N and v than for the same mass of less dense things. And where in Einstein's equation are these two values most potently used? In the numerator. Thus less dense things should have more heat capacity! So the wise teacher selects something cheap and really dense (Pb), some other common and cheap metal that is of medium density (Fe), another cheap and non-dense metal such as aluminum, and forgets about all the other more expensive metals such as brass, copper, bronze, gold, silver and platinum. (Do you hear your pocketbook sighing in relief?) Finally you take a few liquids you have around the lab, and don't forget those two "liquids" - glass and wax.

Okay, here is the way I did it using several different substances. All went rather smoothly: I had on hand a cooler containing ice cubes floating in water into which I had placed all my "substances" so that they would start out as 0°c. Also I had a bucket of water at room temperature.

Some water was scooped out of the bucket with a 20 oz styrofoam cup (5 gm) and weighed. Then I transferred one of the "substances" into the cup, took a new weight, and then swirled it, taking temperatures until it had come to equilibrium. Finally, I dumped everything out and weighed the cup. My data below are already corrected for the weight of the cups used.

Here are my data and calculated specific heat capacities.
(Click for a work sheet for your students.)

Substance Δtemp_water mass_water Δtemp_x mass_x C_rel
ice water 5.8 165 25 40 0.96 ≅ 1.0
lead 2.5 218 28 581 0.034
iron 3.4 165 26.8 107 0.20
glass 2.5 149 27.5 57 0.24 (quartz = 0.22-0.25)
paraffin wax 2.6 141 27 34 0.47
plastic 2.4 178 27 28 0.57

Substance	Δtemp_water	mass_water	Δtemp_x	mass_x	C_rel
ice water	5.8	165	25	40	0.96 ≅ 1.0
lead	2.5	218	28	581	0.034
iron	3.4	165	26.8	107	0.20
glass	2.5	149	27.5	57	0.24 (quartz = 0.22-0.25)
paraffin wax	2.6	141	27	34	0.47
plastic	2.4	178	27	28	0.57

As you can see the cold water into the warmer water gave a C_rel = 1, which is expected and is the CONTROL.

The runs made after water show that the less dense a substance is, the more heat capacity it contains - probably counter to what most people might think, but wise teachers already knew this, because they studied Einstein's equation, above!

In retrospect, the weakest common link in the precision of this experiment is the taking of temperatures as most common lab thermometers do not lend themselves well to reading fractional degrees. Thus if you are unlike most teachers and have an unlimited budget, you might want to purchase a few thermometers with ranges of 0°C to 30°C with intermediary graduations.

SAFETY! I did not use anything of high temperatures because students will undoubtedly burn themselves. Thus I did not put hot objects into cold water, but rather cold objects into room temperature water. Another reason for not using hot water is for the practical reason that water of any temperature other than room temperature will move toward room temperature on its own. The further the temperature is away from room temperature, the faster that movement will be - even without the addition of other objects to the water.

Now let's take a look at how ice (solid water) works in our figuring:

Substance Δtemp_water mass_water Δtemp_ice mass_ice C_rel
ice 5.3 85 23.5 4 4.8 !!!! (how can water be more than water?)

Substance	Δtemp_water	mass_water	Δtemp_ice	mass_ice	C_rel
ice	5.3	85	23.5	4	4.8 !!!! (how can water be more than water?)

There must be something more to this. It's called the heat of fusion (melting) of water ("H_f").
The law of the conservation of energy tells us that:
amt heat lost = amt heat gained
(Δtemp_water) x (mass_water) = the heat energy used in getting 0°C ice to the equilibrium temperature.
The right side can be restated:
(Δtemp_water) x (mass_water) = (used in moving the 0°C water to the equilibrium temperature)
+ (used in changing the 0°C ice into 0°C water)
Filling everything out we get:
(Δtemp_water) x (mass_water) = ((Δtemp_{cold water}) x (mass_{cold water})) + ( x mass_{cold water})
Now solve for H_f and you have the heat of fusion of water-ice!
85 x 5.3 = (4 x 23.5) + 4 H_f
H_f = ((85 x 5.3) - (4 x 23.5))/4 = 89
Because another weak link in this experiment is the weighing of things to the nearest gram, the "4" grams of ice could actually be anywhere from "3.5" to "4.5" grams. The answers for H_f using these are, respectively, 105.2 and 76.6. The real answer is 79.72 cal/gram. So we are within the right range, meaning our way of doing all this is correct; we just need more precision, which can be attained by using better thermometers and a better balance. A discussion might follow for the class on how to improve the precision using the equipment at hand. Answers should include using larger masses (minimizes weighing errors), and, perhaps, using higher temperatures instead of room temperature water. OR, conversely, using room temperature water but having the added items in boiling water (of course that makes a mess with the wax, but you could always pour liquid wax into the water!).

Here is a tricky group quiz question that can be asked. If you added 50 gm of NaBr crystals at 10°C to 250 ml of water at 30°C Assume that you obtain a temperature of 20°C. What is WRONG with this procedure with regard to the determination of the C_rel of NaBr?

Answer (part one): if you had added the NaBr at 30°C rather than at 10°C, the equilibrium temperature would become 40°C - BECAUSE we are more than merely adding a solid to the water - we are DISSOLVING a solid into the water and we must deal with heats of solution. (While dissolving most salts in water cools the water, NaBr however warms the water - just like adding sulfuric acid to water heats it.)

A Quicker Way for a Classroom of Students to Do This

Pre-weigh all of the articles you will place in the ice/water. Have the weights written on the articles - you may find that inscribing (scratching) the numbers onto the articles is better than using marking pen, which might come off in the cold water.

Instead of the students' needing balances to weigh the amount of water they put into their styrofoam cups, you could measure it out. However, the use of graduated cylinders is slow and subject to gaining heat. Thus, you might wish to use metric measuring cups and metric measuring spoons which are available from a number of domestic sources such as Tupperware. For example, transfer of a brimming 50 ml cup of room temperature water is quick and easy. Later, when the article to be tested is ice-water, the use of a 10 ml measuring spoon would work similarly quickly (but don't forget to leave the measuring spoon in the ice-water!).

Precision consideration: You can use this somewhat imprecise method to teach precision. The various groups of students will come up with ranges of results. Ask them what they might do to make that range smaller by using different technology. Answers might be to use more precise thermometers, standardize their thermometers, weigh the water rather than use measuring cups, and increase the amount of temperature change so that it is halfway between ice-water and room temperature water. This could be done increasing or decreasing the masses of the articles added to the room temperature water. There are other ways to improve precision also.

There must be something more to this. It's called the heat of fusion (melting) of water ("H_f"). The law of the conservation of energy tells us that:
amt heat lost =	amt heat gained
(Δtemp_water) x (mass_water) =	the heat energy used in getting 0°C ice to the equilibrium temperature.
The right side can be restated:
(Δtemp_water) x (mass_water) =	(used in moving the 0°C water to the equilibrium temperature) + (used in changing the 0°C ice into 0°C water)
Filling everything out we get:
(Δtemp_water) x (mass_water) =	((Δtemp_{cold water}) x (mass_{cold water})) + ( x mass_{cold water})
Now solve for H_f and you have the heat of fusion of water-ice!
85 x 5.3 =	(4 x 23.5) + 4 H_f
H_f =	((85 x 5.3) - (4 x 23.5))/4 = 89
Because another weak link in this experiment is the weighing of things to the nearest gram, the "4" grams of ice could actually be anywhere from "3.5" to "4.5" grams. The answers for H_f using these are, respectively, 105.2 and 76.6. The real answer is 79.72 cal/gram. So we are within the right range, meaning our way of doing all this is correct; we just need more precision, which can be attained by using better thermometers and a better balance. A discussion might follow for the class on how to improve the precision using the equipment at hand. Answers should include using larger masses (minimizes weighing errors), and, perhaps, using higher temperatures instead of room temperature water. OR, conversely, using room temperature water but having the added items in boiling water (of course that makes a mess with the wax, but you could always pour liquid wax into the water!).