COFFEE CUP CALORIMETERGary L. Bertrand

A coffee cup calorimeter is simply a styrofoam cup (or maybe one cup inside another) to provide insulation when materials are mixed inside of it. A styrofoam cover and a sensitive thermometer complete the apparatus. A more sophisticated apparatus would have a Dewar flask (a double-walled container with reduced pressure between the walls) in a controlled-temperature environment.The most basic experiment involves two of these cups with a measured amount (W_{1}) of hot water at temperature T_{1}in one cup and a measured amount (W_{2}) of cold water at temperature T_{2}in the other cup. The temperatures and weights of the liquids are recorded, then the two are mixed in one of the cups and the final temperature T_{f}is observed.A simple interpretation of this process is that the hot water transfers heat to the cold water, and they both come to the same temperature. If the specific heat of water is c (J/K-g, Joules per Kelvin per gram), the heat (Q_{1}) transferred from the hot water is:Q_{1}= cW_{1}(T_{f}- T_{1}) ,and the heat (Q_{2}) received by the cold water isQ_{2}= cW_{2}(T_{f}- T_{2}) .The insulation of the styrofoam cups prevent any other heat (of course the insulation is not perfect), so the two heats must add to zero:Q_{1}+ Q_{2}= 0 .We can then estimate what the final temperature will be by substituting in this equation:cW_{1}(T_{f}- T_{1}) + cW_{2}(T_{f}- T_{2}) = 0.We can eliminate the specific heat (C) since it is the same for both the hot water and the cold water, and solve for T_{f}:T_{f}= (W_{1}T_{1}+ W_{2}T_{2})/(W_{1}+ W_{2}) .While it is always best to work with absolute temperatures (T,K) rather than Centigrade (t,^{o}C) or Fahrenheit (^{o}F) temperatures, that is not necessary in this simple case (it isn't necessary here because we are working with differences in temperature).

As an example, suppose we mixed 50.0 grams of hot water at 60.0^{o}C with 100.0 grams of cold water at 30.0^{o}C.

From the equation above, we calculate:t_{f}= (50.0 x 60.0 + 100.0 x 30.0)/(50.0 + 100.0)t_{f }= (3000 + 3000)/(150.0)t_{f }= 40.0^{o}C .

Actually, a slightly different final temperature would be observed, probably differing from 40.0^{o}C by 0.2 or 0.3^{o}C. Part of the discrepancy would be due to the imperfect insulation, but some of it would be due to the heat capacity of the cup and the thermometer in which the liquids are mixed. These will also undergo a change from their initial temperature to the final temperature, and this will involve a small amount of heat.

More Advanced Applications:

While there is nothing wrong with doing these calculations in terms of heat, it is important to remember that the heat associated with a change depends on how one gets from the initial condition to the final condition. Changes in thermodynamic properties like energy and enthalpy, however, depend only on the initial and final conditions. For constant pressure processes such as those occurring in these cups which are open to the atmosphere, the enthalpy change is equal to the heat. Under these conditions, the heat capacity is specifically defined as Cp, the heat capacity at constant pressure.

Example (Physical Change - Level 2): 10.0 grams of ice at 0.0^{o}C is added to a coffee-cup-calorimeter containing 100.0 grams of water at 30.0^{o}C. The heat capacity of the apparatus is 10.0 J/K. Calculate the final temperature of the water in the cup.

Solution: It is clear that at least some of the ice is going to melt. There are two possible outcomes: either the final temperature is 0.0^{o}C with some ice remaining, or all of the ice melts and the final temperature cannot be less than 0.0^{o}C. We will assume that all of the ice melts. If this leads to a final temperature which is less than 0.0^{o}C, we will know that we have made a wrong assumption. In that case we would assume that some ice remains and the final temperature is 0.0^{o}C.

Assuming that all of the ice melts, the final temperature will be X^{o}C.

3340 + 41.8 X + 418(X - 30.0) + 10.0(X - 30.0) = 0

(41.8 + 418 + 10.0) X = (418)(30.0) + (10.0)(30.0) - 3340

417 X = 9500

X = 20.2^{o}C.

This is consistent with the assumption that all of the ice melts.

Example (Chemical Change - Neutralization): 200.0 ml of a 1.50 M solution of dichloroacetic acid at 22.40^{o}C is mixed with 200.0 ml of 2.00 M sodium hydroxide at 22.40^{o}C. The heat capacity of the apparatus is 8.6 J/K. The temperature rises to 33.73^{o}C.

(a) How much heat must be transferred to return the final solution to 22.40^{o}C?

Solution: We now have approximately 400 ml of an aqueous solution of neutralized dichloroacetic acid (sodium dichloroacetate) and excess sodium hydroxide at 33.73^{o}C, which must be cooled (along with the apparatus) to 22.40^{o}C.

As a first approximation, we could take the specific heat of the solution to be the same as water and the density of the solution as 1.0 g/ml, so the heat capacity of the solution would be 4.18 J/ml-K. A more accurate approximation is 4.10 J/ml-K.

We have 400 ml of solution and the apparatus at 33.73^{o}C, which must be cooled to 22.40^{o}C. The amount of heat (Q) which must be transferred is the total heat capacity (C) times the change in temperature:C = (400 ml)(4.10 J/ml-K) + 8.6 J/K = 1649 J/K This enthalpy change is due to the neutralization reaction which occurs on mixing.

(b) What is the molar enthalpy of neutralization (kJ/mole) of dichloroacetic acid?

Solution: The amounts (moles) of acid and base are calculated as the product of volume (liters) and concentration (molarity, mole/liter):

amount of acid = (0.2000 L)(1.50 moles/L) = 0.300 molesamount of base = (0.2000 L)(2.00 moles/L) = 0.400 moles

The acid is thelimiting reagent, so 0.300 moles of acid and base are neutralized, and 0.100 moles of hydroxide ion remain.

The enthalpy change for the neutralization of 0.300 moles of acid is - 18.67 kJ, so themolarenthalpy of neutralization of dichloroacetic acid is