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Heat capacity problems — solutions

This will cover the solutions to the three complicated heat capacity questions.

These worked answers will go through the previous problems in depth.

Calculation 1

The heat capacity of water is 4.18 J g-1 K-1. Calculate the energy, in kJ, required to boil 1.2 L of water, starting at 25 °C.
Boiling water from 25 °C requires raising its temperature by another 75 °C, so ΔT = 75 K (100-75).
The mass of water in g, is calculated by multiplying volume by density. This is thankfully very easy for water because it’s defined (at room temperature) as 1 g cm-1. If we have 1 L, there are 1000 mL, and therefore 1000 g. The energy is therefore:
4.18 J g-1 K-1 × 1000 g × 75 K = 313500 J = 313.5 kJ

Calculation 2

A sample of 150 mL water of temperature T1=25 °C is added to a sample of 50 mL water of temperature T2=60 °C in a foam cup calorimeter. What is the final temperature, T0, of the system when thermal equilibrium is reached? Answer in °C.
A weighted average of these temperatures will get you the right answer:
  • 150/200 * 25 + 50/200 * 60 = 33.75 °C
For a more thermodynamically robust approach, we should consider the heat each solution could contribute.
So, we can calculate the heat released by each solution to get both to the same temperature (e.g., 0 °C for convenience, as ΔT is just the temperature in °C), and then add that energy to the combined solution at 0°, and predict the temperature change. This is not too dissimilar to Hess’ law, or conservation of energy.
  • 150 g of water at 25 °C gives: 4.18 J g-1 K-1 × 150 g × 25 K = 15.6 kJ
  • 50 g of water at 60 °C gives: 4.18 J g-1 K-1 × 50 g × 60 K = 12.5 kJ
Note that there’s slightly less energy in the hotter solution because there is less of it.
So a total energy available of 15.6 + 12.5 = 28.1 kJ.
150 g + 50 g = 200 g. So if 28.1 kJ of energy is given to 200 g of water at 0 °C, the final temperature is:
  • 28215 J / (200 g × 4.18 J K-1 g-1) = 33.75 °C

Calculation 3

50.0 mL 0.250 M HCl at 19.50 °C is mixed with 50.0 mL of 0.250 M NaOH at 19.50 °C in a calorimeter. After mixing, solution temperature rises to 21.21 °C. The heat capacity of the resulting salt solution is 4.18 J g-1 K-1. Assume a density of 1 g cm-3. Calculate the heat of this reaction in kJ mol-1.
The total volume is 100 mL, therefore 100 g of liquid.
The temperature increase is 21.21 – 19.5 = 1.71 K. (do not add 273 to this number!)
So the energy released is:
  • -4.18 J K-1 g-1 × 1.71 K × 100 g = -715 J
There are 0.25 mol dm-3 × 0.05 dm3 = 0.0125 mol of reactions occurring. (This is important to realise it’s not 0.0125 moles of NaOH + 0.0125 moles of HCl to equal 0.025 mol!)
Therefore, the energy released per mole is:
  • -715 J / 0.0125 mol = -57,200 J mol-1
Or, -57.2 kJ mol-1.
This value is negative because it represents energy lost by the system (the reaction).

Calculation 4

A reaction releases 64 kJ mol-1 of heat, into 620 mL of water (1 g cm-3), which has a heat capacity of 4.18 J g-1 K-1. There are 0.19 moles of reactants in the container. Predict the subsequent temperature increase.
The key observation here is that it’s the reverse of how this is done experimentally – you’re predicting a temperature increase from a known enthalpy. You will also need to be careful of matching the moles and masses. The heat given off is in per mole, so multiply by the number of moles to get the total energy in J or kJ.
  • 0.19 mol × 64 kJ mol-1 = 12.16 kJ = 12,160 J
The energy change of the reaction is normally given by:

[q = -C_p times m times Delta T]

Which can be rearranged to get the temperature increase:

[frac{-q}{C_ptimes m}=Delta T]

You’re warming up the solution, so the heat capacity is the one given, and the mass is the volume of solvent – assuming 1 g cm-3, the mass in g is the same as the volume in cm3. In this example, 620 g.
  • 12,160 J / (4.18 J K-1 g-1 × 620 g) = +4.7 K
In reality, the heat needs to heat up the solvent and any reactants, and also heat up the calorimeter. So the heat capacity of solution will be slightly different to the value given here, and dependent on concentration and the identity of the reactants and products. This isn’t included in the question above, so you can ignore it for this example.

Calculation 5

A 1.5 g sample of NH4NO3 (s) is added to 35.0 g of water in a foam cup and stirred until it dissolves. The temperature of the solution drops from 22.7 to 19.4 C. What is the heat of dissolution of NH4NO3, expressed in kJ mol-1. Specific heat capacity of the total solution is 4.18 J K-1 g-1.

One key observation is is the total mass. 35 g of water is the solvent, and 1.5 g is added to it. The total solution at the end must weight 36.5 g! This is important because the energy must exchange heat with everything in the system that we’re looking at, which includes the material dissolved in the water.

The temperature difference is -3.3 K. Therefore, the energy is:

  • -4.18 J K-1 g-1 × 36.5 g × -3.3 K = +503 J

The enthalpy change is positive because the temperature has gone down. This means energy has been taken from the surroundings (where we measure temperature) and into the system. The enthalpy of the system, therefore, has increased.

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Introduction to Thermodynamics

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