Interactive Thermodynamics Laboratory
Calorimetry
Calorimetry is the experimental measurement of heat exchange during physical or chemical processes. Based on the Law of Conservation of Energy, the heat lost by hot materials must equal the heat gained by cooler materials in a closed system. Use the simulator tabs below to drop hot metals into insulated cups, trace temperature curves, and solve for specific heat capacities.
Calorimetry Simulation Laboratory
What is Calorimetry?
In thermodynamics, calorimetry is the experimental science of measuring the quantity of thermal energy transferred as heat during physical or chemical processes. It is used to determine values such as specific heat capacities of unknown substances, enthalpies of chemical reactions, and latent heats of phase changes.
The core principle rests directly upon the Law of Conservation of Energy. Inside a closed, perfectly insulated container (a calorimeter) where no thermal energy can enter or escape to the surroundings, any thermal energy lost by hot components must be completely absorbed by cold components. We write this balance as:
mm × cm × (Tm - Tf) = mw × cw × (Tf - Tw)
mm, mw = Mass of metal block and water (kg)
cm, cw = Specific heat capacity of metal and water (J/kg·K)
Tm, Tw = Initial temperature of metal block and water (°C)
Tf = Final equilibrium temperature of the mixture (°C)
How a Calorimeter Works
A basic calorimeter (such as the standard nested polystyrene "coffee-cup" calorimeter used in physics classrooms) consists of several critical parts designed to maintain a closed thermal system:
- Double-walled insulation: Air spaces or low-conductivity walls prevent heat conduction to the outside air.
- Tight lid: Prevents heat loss via convection currents and evaporation of liquid vapor.
- Thermometer: Reads the temperature rise or drop of the mixture.
- Stirrer: A metal or plastic loop wiggled up and down to distribute heat evenly, ensuring the mixture reaches thermal equilibrium quickly.
Specific Heat Capacities Table
Materials react differently to heat energy. Materials with low specific heat capacities (like metals) experience rapid temperature swings compared to water, which has a very high specific heat capacity:
| Substance | Specific Heat Capacity (c) (J/kg·K) | Relative Thermal Response |
|---|---|---|
| Liquid Water | 4,184 | Extremely slow temperature change (Standard benchmark) |
| Aluminium | 900 | Moderate temperature change |
| Iron / Steel | 450 | Fast temperature change |
| Copper | 385 | Very fast temperature change |
| Lead | 128 | Extremely rapid temperature change |
Constant-Pressure Calorimetry
Commonly used in solutions (coffee-cup). The pressure remains atmospheric, meaning heat exchange represents enthalpy change (ΔH).
Constant-Volume (Bomb) Calorimetry
A rigid steel container ("bomb") placed in water. Used for highly explosive or combustion reactions. Volume is constant, measuring internal energy change (ΔU).
Solved Examples
Solved Example 1: A 0.15 kg copper block at 100°C is dropped into an insulated cup containing 0.25 kg of water at 20°C. If the heat capacity of the cup is ignored, calculate the final equilibrium temperature. (c_copper = 385 J/kg·K, c_water = 4,184 J/kg·K)
Identify the given values for the metal block: m_m = 0.15 kg, c_m = 385 J/kg·K, T_m = 100°C.
Identify the given values for the water: m_w = 0.25 kg, c_w = 4,184 J/kg·K, T_w = 20°C.
State the calorimetry equation: Q_lost = Q_gained => m_m * c_m * (T_m - T_eq) = m_w * c_w * (T_eq - T_w).
Substitute values and expand: 0.15 * 385 * (100 - T_eq) = 0.25 * 4,184 * (T_eq - 20).
Simplify coefficients: 57.75 * (100 - T_eq) = 1,046 * (T_eq - 20) => 5,775 - 57.75 * T_eq = 1,046 * T_eq - 20,920.
Solve for T_eq: 5,775 + 20,920 = (1,046 + 57.75) * T_eq => 26,695 = 1,103.75 * T_eq => T_eq ≈ 24.19°C.
Verify: The final temperature (24.19°C) is between the initial hot metal (100°C) and initial cool water (20°C) temperatures.
Solved Example 2: A 0.20 kg sample of an unknown metal heated to 95°C is dropped into 0.15 kg of water at 15°C inside a calorimeter. The final equilibrium temperature is measured to be 22.4°C. Neglecting heat losses to the cup, calculate the specific heat capacity of the metal.
Identify the given parameters: m_metal = 0.20 kg, T_metal = 95°C, m_water = 0.15 kg, T_water = 15°C, T_equilibrium = 22.4°C.
State the heat balance: Q_lost = Q_gained => m_metal * c_metal * (T_metal - T_eq) = m_water * c_water * (T_eq - T_water).
Substitute known parameters: 0.20 * c_metal * (95 - 22.4) = 0.15 * 4,184 * (22.4 - 15).
Simplify: 0.20 * c_metal * 72.6 = 627.6 * 7.4 => 14.52 * c_metal = 4,644.24.
Calculate c_metal: c_metal = 4,644.24 / 14.52 ≈ 319.85 J/kg·K.
Verify: The value (approx. 320 J/kg·K) is typical for heavy metallic elements such as gold, lead, or copper alloys.
Solved Example 3: A 0.10 kg gold block (c = 129 J/kg·K) at 120°C is dropped into a glass calorimeter cup of mass 0.05 kg containing 0.20 kg of water at 18°C. If the specific heat of glass is 840 J/kg·K, find the equilibrium temperature.
Identify the system elements: gold block (m_m = 0.10 kg, c_m = 129 J/kg·K, T_m = 120°C), water (m_w = 0.20 kg, c_w = 4,184 J/kg·K, T_w = 18°C), and glass cup (m_cup = 0.05 kg, c_cup = 840 J/kg·K, T_cup = 18°C).
State the calorimetry equation including the cup: Q_lost = Q_gained => m_m * c_m * (T_m - T_eq) = (m_w * c_w + m_cup * c_cup) * (T_eq - T_w).
Substitute values: 0.10 * 129 * (120 - T_eq) = (0.20 * 4,184 + 0.05 * 840) * (T_eq - 18).
Simplify coefficients: 12.9 * (120 - T_eq) = (836.8 + 42) * (T_eq - 18) => 1,548 - 12.9 * T_eq = 878.8 * T_eq - 15,818.4.
Gather terms: 1,548 + 15,818.4 = (878.8 + 12.9) * T_eq => 17,366.4 = 891.7 * T_eq.
Solve for T_eq: T_eq = 17,366.4 / 891.7 ≈ 19.48°C.
Verify: Taking the glass container's thermal mass into account slightly reduced the final temperature compared to a hypothetical perfectly insulated massless cup.
Summary of Key Concepts
- Thermal Equilibrium occurs when heat transfer ceases because all components reach a single final temperature (T_f).
- Q = m × c × ΔT describes sensible heat exchange where c is specific heat capacity.
- For phase transitions, calorimeter cups must also account for latent heat (e.g. melting ice: Q = m × L_f).
- Ignored heat capacity of the cup introduces experimental error. A real calorimeter container holds a calibrated "Water Equivalent".
Practice Questions
- What is the Principle of Calorimetry? State the law of physics it is based on.
Show Explanation
The Principle of Calorimetry states that for a closed insulated system, the heat energy lost by hot bodies must equal the heat energy gained by cold bodies until they reach thermal equilibrium: Q_lost = Q_gained. This is based on the Law of Conservation of Energy.
- Why is a calorimeter vessel usually made of copper and insulated with outer layers?
Show Explanation
Copper is used because it has a low specific heat capacity (absorbing very little heat itself) and is an excellent thermal conductor (enabling fast heat sharing to ensure uniform temperature). It is surrounded by insulated outer layers to prevent heat exchange with the laboratory environment, maintaining a closed system.
- Identify the difference between a constant-pressure (coffee-cup) calorimeter and a bomb calorimeter.
Show Explanation
A coffee-cup calorimeter operates under constant atmospheric pressure and is typically used for measuring simple heat changes in liquid solutions. A bomb calorimeter operates under constant volume (sealed steel vessel) and is used for measuring combustion heats of substances ignited in pressurized oxygen.
- During a calorimetry lab, a student does not stir the liquid. Explain how this affects the experimental results.
Show Explanation
Without stirring, heat is not distributed uniformly throughout the liquid. The thermometer will read localized hot or cold pockets, leading to incorrect temperature measurements and yielding inaccurate specific heat calculations.
- A hot metal block is transferred slowly from the heating bath to the calorimeter cup. How does this delay affect the calculated specific heat capacity?
Show Explanation
During the slow transfer, the metal block loses heat to the surrounding air. When dropped into the cup, it is colder than measured, releasing less heat to the water. This results in a lower final temperature, making the calculated specific heat capacity of the metal appear falsely low.
- Why does adding a cover or lid to the calorimeter cup improve experimental accuracy?
Show Explanation
The lid prevents heat loss due to convection and evaporation of the hot water vapor to the outside atmosphere, keeping the system closed.
- A 0.50 kg metal block at 100°C is dropped into 0.50 kg of water at 20°C. If their temperatures meet at 30°C, what is the ratio of their specific heat capacities (c_metal / c_water)?
Show Explanation
Using Q_lost = Q_gained: m_m * c_m * (T_m - T_eq) = m_w * c_w * (T_eq - T_w). Since masses are equal (0.50 kg), this simplifies to: c_m * (100 - 30) = c_w * (30 - 20) => c_m * 70 = c_w * 10 => c_m / c_w = 10 / 70 = 1 / 7 ≈ 0.143.
Frequently Asked Questions
What is calorimetry?
Calorimetry is the science of measuring the quantity of heat energy transferred during physical processes, chemical reactions, or state changes.
What is a calorimeter?
A calorimeter is an insulated laboratory instrument used to measure heat flow. A simple classroom version consists of a nested polystyrene or copper cup with a lid, stirrer, and thermometer.
Why does a calorimeter ignore heat exchange with the air?
We assume the calorimeter is a perfectly closed insulated system where no heat escapes to or enters from the surroundings, allowing us to write Q_lost = Q_gained.
What is the formula used in calorimetry?
The core equation is Q = m × c × ΔT, alongside the conservation statement Q_lost = Q_gained. This expands to mm × cm × (Tm - Tf) = mw × cw × (Tf - Tw).
Why is the stirrer wiggled up and down in a calorimeter?
Stirring mixes the hot and cold water layers, ensuring the system reaches a uniform thermal equilibrium temperature quickly.
Can calorimetry be used to measure latent heat?
Yes. By dropping ice at 0°C into warm water, the heat gained includes both the latent heat of fusion (melting) and sensible warming: Q = m × L_f + m × c × ΔT.
What is the specific heat capacity of water?
The specific heat capacity of water is exceptionally high, approximately 4,184 J/kg·K (or 1.00 cal/g·°C).
What material holds the lowest specific heat capacity in general labs?
Heavy metals like gold (129 J/kg·K) and lead (128 J/kg·K) have low specific heat capacities, meaning their temperatures change rapidly when heat is added.
Does the water equivalent of the calorimeter matter?
Yes. The calorimeter cup itself absorbs some heat. The "water equivalent" is the mass of water that would absorb the same amount of heat as the cup for the same temperature change.
Is calorimetry an endothermic or exothermic measurement?
It measures both. An endothermic reaction will lower the calorimeter water temperature, while an exothermic reaction will raise it.