Calorimetry

5.2 Calorimetry

Background: The System and the Surroundings

Calorimetry is the technique used to measure the amount of heat transferred to or from a substance during a chemical or physical process. To understand the math, we must define two areas:

The System: The specific substance(s) undergoing the change (e.g., the chemical reaction).
The Surroundings: Everything else in contact with the system that serves to provide or absorb heat (e.g., the water in the calorimeter, the cup itself, the thermometer).

A Calorimeter is a device designed to minimize heat exchange with the outside world, effectively trapping the heat so we can measure it. In a "perfect" calorimeter, the heat lost by the system is exactly equal to the heat gained by the surroundings:

q_system + q_surroundings = 0
(or: q_system = −q_surroundings)

✏️ Integrated Practice: Heat Transfer Between Substances

A 360.0-g piece of hot steel rebar is dropped into 425 mL of water (425 g) initially at 24.0 °C. The water temperature rises to a final equilibrium of 42.7 °C.

Step 1 (Part A): Calculating Heat Gained by Water
Using the specific heat of water (4.184 J/g°C), calculate how many Joules of heat the water absorbed from the metal.
Hint: q_w = c_w × m_w × (T_final - T_{initial, water})

Click for Part A Answer

q_w = (4.184 J/g°C) × (425 g) × (42.7°C - 24.0°C)
q_w = 33,250 J

Step 2 (Part B): Determining the Metal's Initial Temperature
Since q_rebar = −q_water, the rebar lost 33,250 J (q = -33,250 J). If the specific heat of steel is 0.449 J/g°C, what was the initial temperature of the rebar?
Question: Rearrange q = c × m × (T_f - T_i) to solve for T_i.

Click for Part B Answer

-33,250 J = (0.449 J/g°C) × (360.0 g) × (42.7°C - T_i)
-33,250 / 161.64 = 42.7 - T_i
-205.7 = 42.7 - T_i
T_i = 248.4 °C

Background: Reaction Calorimetry

When we perform a chemical reaction inside a calorimeter, the aqueous solution absorbs or releases the heat of the reaction. For these problems, we often assume the solution has the same density (1.0 g/mL) and specific heat (4.184 J/g°C) as pure water.

✏️ Integrated Practice: Exothermic Neutralization

50.0 mL of HCl and 50.0 mL of NaOH are mixed (Total volume = 100.0 mL). The temperature of this 100.0 g "surroundings" solution rises from 22.0 °C to 28.9 °C.

Step 1 (Part A): Calculating Solution Heat
Calculate the heat gained by the 100.0 g solution (q_soln) using the specific heat of water. Click for Part A Answer

q_soln = (4.184 J/g°C) × (100.0 g) × (28.9°C - 22.0°C)
q_soln = 2,887 J (approx. 2.9 kJ)

Step 2 (Part B): Interpreting the Reaction
Based on your answer in Part A, is the chemical reaction exothermic or endothermic, and what is the value of q_rxn? Click for Part B Answer

Since the solution gained heat, the reaction must have released it.
q_rxn = −2.9 kJ. This is an Exothermic reaction.

Clinical Application: Hand Warmers

Exothermic reactions are used in medical hand warmers. Some use the crystallization of sodium acetate, while others use the oxidation of iron. In both cases, the chemical system releases heat that your hands (the surroundings) absorb.