D. Energy Changes and Rates of Reaction · Energy Changes and Rates of Reaction

Heat Transfer and Coffee-Cup Calorimetry

D3.3 D2.3 D2.4 D2.1 A1.5 A1.8 A1.13About 75 minutes

Learning goals

I can explain how mass, specific heat capacity and temperature change determine the heat gained or lost by a substance.
I can use q = mcΔT, with c in J/(g·°C), to calculate heat transfer.
I can use coffee-cup calorimetry data to calculate a molar enthalpy of solution or neutralization with the correct sign.
I can state the assumptions of simple calorimetry and calculate percentage error.

Words to know

specific heat capacity (c)
The energy needed to raise the temperature of 1 g of a substance by 1 °C, in J/(g·°C). Water: 4.18 J/(g·°C).

heat capacity (C)
The energy needed to raise the temperature of a whole object by 1 °C, in J/°C. C = mc.

calorimetry
Measuring the heat released or absorbed by a change from the temperature change of its surroundings, usually water.

calorimeter
An insulated container in which calorimetry is done.

molar enthalpy of solution (ΔH_soln)
The enthalpy change when one mole of a substance dissolves in water.

percentage error
(experimental value − accepted value) ÷ accepted value × 100%.

Measuring heat you cannot see

You cannot hold a thermometer inside a molecule to watch its bonds change. But you can let a reaction happen in water and watch the water's temperature. Water is the perfect "energy accountant": it is cheap, safe, and it has a large, well-known specific heat capacity. This lesson shows you how to turn a temperature change into an enthalpy change. The technique is called calorimetry, and it is how most of the numbers in thermochemistry tables were first measured.

Q = mcΔT in chemistry units

Remember from Grade 11 Physics (SPH3U): the heat gained or lost by a substance is

q = mcΔT

q = heat (J). (Physics wrote Q; chemists usually write a lower-case q.)
m = mass. In chemistry we use grams.
c = specific heat capacity, in J/(g·°C): the energy needed to raise 1 g by 1 °C. For water, c = 4.18 J/(g·°C). (This is the same as the 4.18 × 10³ J/(kg·°C) you used in physics, just per gram.)
ΔT = T_final − T_initial (°C). Positive when the substance warms.

The equation explains how three things control the heat (expectation D3.3):

Mass: twice the mass needs twice the heat for the same temperature change.
Specific heat capacity: a substance with a larger c (water, 4.18) needs more energy per degree than one with a small c (copper, 0.385). That is why a metal spoon heats quickly but a pot of soup takes a while.
Temperature change: the heat is proportional to ΔT.

Sometimes we know a whole object's heat capacity, C = mc, in J/°C. Then q = CΔT.

Substance	c (J/(g·°C))
water (liquid)	4.18
ethanol (liquid)	2.44
aluminum	0.897
glass (typical)	0.84
iron	0.449
copper	0.385

Specific heat capacities near room temperature. Water's is unusually large.

The key idea: the system's loss is the surroundings' gain

In a calorimeter, the reaction (the system) happens inside water (the surroundings). If no heat escapes to the outside:

q_system = −q_surroundings

So if the water absorbs +2.80 kJ (its temperature rises), the reaction released 2.80 kJ, and its ΔH is negative. If the water loses energy (temperature falls), the reaction absorbed it, and ΔH is positive. Finally, divide by the amount that reacted to get a molar enthalpy:

ΔH = q_system ÷ n = −(m c ΔT)_water ÷ n

A coffee-cup calorimeter: two nested polystyrene cups with a lid, a thermometer and a stirrer. It works at constant (atmospheric) pressure, so the heat measured is the enthalpy change.

The coffee-cup calorimeter and its assumptions

A simple coffee-cup calorimeter is used for reactions in solution: neutralizations, dissolving solids, metals reacting with acids. To keep calculations manageable we assume:

All the heat goes into (or comes from) the water: none is lost to the cup, the thermometer or the air.
A dilute aqueous solution has the same density (1.00 g/mL) and specific heat capacity (4.18 J/(g·°C)) as pure water.
The mass of a dissolved solid is usually added to the mass of water, but small masses are sometimes ignored. Read each question carefully.

Real calorimeters do lose some heat. A good way to correct for slow heat loss is to record the temperature every 30 seconds before and after mixing, graph it, and extrapolate the slow cooling line back to the moment of mixing.

Sample data for a neutralization. The temperature peaks at about 27.5 °C and then slowly falls as heat escapes. Extending the cooling line back to the moment of mixing gives a corrected maximum (about 27.9 °C) that accounts for the heat lost.

Worked example

Example 1: Heat absorbed by water

How much heat is needed to warm 250.0 g of water from 18.5 °C to 65.0 °C?

ΔT = 65.0 − 18.5 = 46.5 °C.
q = mcΔT = 250.0 g × 4.18 J/(g·°C) × 46.5 °C = 48 592 J.
Round to three significant digits (4.18 and 46.5 each have three).

q = 4.86 × 10⁴ J = 48.6 kJ absorbed by the water.

Worked example

Example 2: Molar enthalpy of neutralization

50.0 mL of 1.00 mol/L HCl(aq) and 50.0 mL of 1.00 mol/L NaOH(aq), both at 21.00 °C, are mixed in a coffee-cup calorimeter. The highest temperature is 27.70 °C. Find the molar enthalpy of neutralization of HCl.

Mass of solution: 100.0 mL × 1.00 g/mL = 100.0 g. ΔT = 27.70 − 21.00 = 6.70 °C.
q_water = mcΔT = 100.0 g × 4.18 J/(g·°C) × 6.70 °C = 2801 J = 2.80 kJ (absorbed).
So q_reaction = −2.80 kJ (released).
n(HCl) = cV = 1.00 mol/L × 0.0500 L = 0.0500 mol. (NaOH is also 0.0500 mol; neither is in excess.)
ΔH_neut = −2.80 kJ ÷ 0.0500 mol = −56.0 kJ/mol.

ΔH_neut = −56.0 kJ/mol of HCl. The accepted value for a strong acid with a strong base is about −56 to −57 kJ/mol, because the reaction is always H⁺(aq) + OH⁻(aq) → H₂O(l).

Worked example

Example 3: An endothermic heat of solution

8.00 g of ammonium nitrate, NH₄NO₃ (M = 80.06 g/mol), dissolves in 100.0 g of water in a calorimeter. The temperature falls from 22.00 °C to 16.40 °C. Find ΔH_soln and the percentage error, if the accepted value is +25.7 kJ/mol.

Mass of solution = 100.0 + 8.00 = 108.0 g. ΔT = 16.40 − 22.00 = −5.60 °C.
q_solution = 108.0 g × 4.18 J/(g·°C) × (−5.60 °C) = −2528 J. The water lost 2.528 kJ.
q_system = +2.528 kJ (absorbed by the dissolving).
n = 8.00 g ÷ 80.06 g/mol = 0.09993 mol.
ΔH_soln = +2.528 kJ ÷ 0.09993 mol = +25.3 kJ/mol.
Percentage error = (experimental − accepted) ÷ accepted × 100% = (25.3 − 25.7) ÷ 25.7 × 100% = −1.6%.

ΔH_soln = +25.3 kJ/mol (endothermic), a −1.6% error. The experimental value is slightly low in size, which fits some heat leaking into the cold cup from the warmer room.

Worked example

Example 4: Predicting a final temperature

5.00 g of NaOH(s) dissolves in 100.0 g of water at 20.0 °C. ΔH_soln = −44.5 kJ/mol (M = 40.00 g/mol). Predict the final temperature.

n = 5.00 ÷ 40.00 = 0.125 mol. Heat released = 0.125 mol × 44.5 kJ/mol = 5.56 kJ = 5563 J.
The solution (105.0 g) absorbs it: ΔT = q ÷ (mc) = 5563 J ÷ (105.0 g × 4.18 J/(g·°C)) = 12.7 °C.
T_final = 20.0 + 12.7 = 32.7 °C.

About 32.7 °C. (Solid NaOH is corrosive; in a real lab you would wear goggles and gloves.)

Hands-on activity

Build and test a foam-cup calorimeter

You need: two polystyrene (foam) cups that nest, a plastic lid or a piece of cardboard with a hole, a kitchen thermometer, a measuring cup, a kitchen scale if you have one, hot tap water, cold tap water, a spoon, and a grown-up. Safety: use hot tap water, not boiling water, and let a grown-up pour it.

Part A: how good is your calorimeter?

Nest the cups. Measure 100 mL (100 g) of cold water into the calorimeter and record its temperature.
Measure 100 mL of hot tap water in a separate cup and record its temperature.
Pour the hot water into the calorimeter, cover, stir, and record the highest steady temperature.
Predict the final temperature if no heat were lost: equal masses of water end up at the average of the two starting temperatures. Compare.

Part B: an endothermic reaction

Put 50 mL of vinegar in the calorimeter and record its temperature every 15 s for 1 minute.
Add 1 level teaspoon of baking soda (do it over the sink, it foams), cover loosely, stir and keep recording for 3 minutes.
Calculate q for the vinegar (assume 50 g and c = 4.18 J/(g·°C)) and state the sign of ΔH for the reaction.

What to look for: in Part A the measured temperature is usually 1 to 3 °C below the average, because the cups, thermometer and air absorbed some heat. In Part B the temperature drops by a few degrees: the reaction is endothermic, so q_vinegar is negative and ΔH is positive. Some energy also leaves with the escaping carbon dioxide foam, so this is a rough measurement.

Just for fun

Why did the chemist bring two foam cups to the party?

She heard it was going to be a hot event and wanted to keep things insulated, but not isolated. She is still a very social system.

Your turn

How much heat is released when 500.0 g of water cools from 80.0 °C to 25.0 °C?

A 25.0 g piece of aluminum absorbs 900. J. By how much does its temperature rise? (c = 0.897 J/(g·°C))

In a calorimeter, the water temperature rises by 3.20 °C. Is the reaction exothermic or endothermic, and what is the sign of ΔH?

100.0 mL of 0.500 mol/L HCl and 100.0 mL of 0.500 mol/L NaOH, both at 20.50 °C, are mixed. The final temperature is 23.85 °C. Find ΔH_neut per mole of HCl.

2.50 g of KCl (M = 74.55 g/mol) dissolves in 50.0 g of water. The temperature falls from 21.40 °C to 18.60 °C. Find ΔH_soln. (Use the solution mass, 52.5 g.)

Explain how mass, specific heat capacity and temperature change each affect the heat absorbed by a substance.

State two assumptions made in coffee-cup calorimetry.

More practice: warm-up

What is the specific heat capacity of water in J/(g·°C)?

Write the equation that relates heat to mass, specific heat capacity and temperature change.

What mass is 75.0 mL of a dilute aqueous solution, using the usual assumption?

If the water in a calorimeter gains 1.50 kJ, how much heat did the reaction release?

Why are two nested foam cups used instead of one glass beaker?

More practice: on your own

A 40.0 g block of iron at 95.0 °C is dropped into 100.0 g of water at 20.0 °C. Find the final temperature. (c_iron = 0.449 J/(g·°C))

Dissolving 4.00 g of CaCl₂ (M = 110.98 g/mol) in 100.0 g of water raises the temperature from 20.00 °C to 26.40 °C. Find ΔH_soln (use 104.0 g of solution).

Predict the temperature change when 0.0500 mol of HNO₃ is neutralized by excess NaOH in a total of 150.0 g of solution. (ΔH_neut = −56.0 kJ/mol)

A student finds ΔH_neut = −49 kJ/mol; the accepted value is −56 kJ/mol. Calculate the percentage error and give one likely cause.

Why does extrapolating the cooling curve back to the time of mixing give a better ΔT than simply reading the highest temperature?

Challenge

A calorimeter (cups, thermometer and stirrer) has a heat capacity of 15.0 J/°C. It contains 100.0 g of water. When 0.0400 mol of a substance dissolves, the temperature rises from 21.00 °C to 25.50 °C. Find ΔH_soln, including the calorimeter's heat.

50.0 mL of 2.00 mol/L HCl is mixed with 75.0 mL of 1.00 mol/L NaOH, both at 20.00 °C. Using ΔH_neut = −56.0 kJ/mol, predict the final temperature. (Identify the limiting reactant first.)

Why do you think the molar enthalpy of neutralization is nearly the same (about −56 kJ/mol) for HCl + NaOH, HNO₃ + KOH and HCl + KOH?

Check your understanding

Quiz

How much heat is needed to raise 200.0 g of water by 5.00 °C?

In a coffee-cup calorimeter the temperature drops from 22.0 °C to 18.0 °C. Which is true?

Which substance needs the most energy to raise 1 g of it by 1 °C?

A reaction releases 2.09 kJ into 100.0 g of water. What is the temperature change of the water?

Which change would most improve the accuracy of a coffee-cup calorimetry experiment?

When 0.0200 mol of a salt dissolves, the water in a calorimeter loses 0.500 kJ. What is ΔH_soln?

For parents and teachers

Big ideas (D3.3, D2.3, D2.4): the heat a substance gains or loses depends on its mass, its specific heat capacity and its temperature change (q = mcΔT). In a calorimeter, the heat released or absorbed by a reaction is measured by the temperature change of the water around it: q_reaction = −q_water. Dividing by moles gives a molar enthalpy.

How to help: Do Part A of the activity together and ask your teen to explain why the result is below the predicted average. Ask: "If the water got warmer, what sign is ΔH?" (negative) until the answer is automatic.

Common difficulties:

Sign errors. The water's q and the reaction's q have opposite signs.
Using the mass of the solid instead of the mass of the solution in q = mcΔT.
Dividing by the wrong number of moles (use the limiting reactant, or the substance the question names).
Mixing J and kJ; convert before dividing.
Units of c: chemistry uses J/(g·°C), physics used J/(kg·°C).

Activity safety: a grown-up pours the hot tap water (never boiling water in foam cups, which can soften). Vinegar and baking soda foam over; do Part B over a sink.