Calorimetry and Heat

Calorimetry measures heat transferred during a process. Drop a hot metal object into cooler water. The metal cools while the water warms. Measuring their temperature changes helps calculate the energy exchanged.

• Heat: energy transferred because systems have different temperatures.
• Temperature: a measure of hotness or coldness.
• Thermal equilibrium: equal temperatures, with no net temperature-driven heat transfer.

Heat and temperature describe different quantities. A thermometer measures temperature, rather than the total energy transferred. Heat is measured in joules, written J. Net heat transfers spontaneously from hotter material toward colder material.

Specific heat describes the energy needed per unit mass for a temperature rise. Its symbol is c. Its SI unit is J/(kg·K).

For sensible heating, use Q = mcΔT. Sensible heating changes temperature without changing phase.

A temperature difference of 1°C equals a difference of 1 K. Do not confuse a temperature difference with an absolute temperature.

Use this formula for fixed mass and effectively constant specific heat. Assume no phase change within the calculation. The simple heating model also excludes other energy inputs, such as mechanical work.

What does the formula predict?

Doubling the mass doubles the heat needed for the same temperature rise. Doubling the temperature rise also doubles the heat needed, when c stays constant.

Liquid water has a representative c of 4186 J/(kg·K) near 15°C. This value is an approximation for calculations. Specific heat can change with material, temperature and physical state.

Worked example: how much heat warms half a kilogram of water?

Take 0.50 kg of liquid water warming from 15°C to 25°C. Use the representative value c = 4186 J/(kg·K) throughout this interval. Assume no phase change or other energy input.

1. Find the temperature change: ΔT = 25 − 15 = 10°C = 10 K.
2. Substitute the values: Q = 0.50 × 4186 × 10.
3. Calculate: Q = 20,930 J ≈ 20.9 kJ.

The positive result means the water gains heat under this model. A temperature rise alone does not always prove heat entered. Mechanical work can also warm a body.

Heat capacity describes the whole object's response. Specific heat describes the material per unit mass. For the stated sample and conditions, C = mc.

C is heat capacity, measured in J/K. Thus, a larger water sample has a larger heat capacity at the same specific heat. Our 0.50 kg sample has C = 0.50 × 4186 = 2093 J/K.

A calorie is another energy unit. The 15°C calorie is approximately 4.186 J. The thermochemical calorie is defined as exactly 4.184 J. Use the definition and units supplied in the problem.

The cooler water and the calorimeter can both gain energy. The calorimeter is the measuring vessel and associated apparatus. Insulation reduces energy exchange with the outside.

External insulation does not remove the vessel's own heat capacity. Even a perfectly externally insulated vessel can absorb heat internally.

Write the balance before calculating

Use positive Q for heat gained and negative Q for heat lost. In an isolated mixing model, the heat transfers sum to zero. Assume no phase changes, reactions or significant work during mixing.

Q_hot + Q_water + Q_cal = 0

Let a hot object start at T_h. Let the cooler water and calorimeter both start at T_c. Let every component reach a common final temperature T_f.

m_h c_h(T_f − T_h) + m_w c_w(T_f − T_c) + C_cal(T_f − T_c) = 0

• m_h and m_w are the object and water masses, in kg.
• c_h and c_w are their specific heats, in J/(kg·K).
• C_cal is the apparatus heat capacity, in J/K.
• Each temperature difference is measured in K, or equivalently °C.

The hot object's term is negative because T_f is below T_h. The water and vessel terms are positive when both warm.

For example, suppose the vessel has C_cal = 100 J/K and warms by 5 K. Its absorbed heat is 100 × 5 = 500 J. Omitting that term leaves 500 J unaccounted for in this hypothetical experiment.

What is water equivalent?

Water equivalent is the water mass with the same heat capacity as the apparatus. Write W = C_cal/c_w, where W is measured in kg. For the same temperature change, W c_w ΔT equals C_cal ΔT.

Water equivalent does not mean equal heat for arbitrary, different temperature changes. If the vessel starts at another temperature, use its own initial temperature in the balance.

During melting, supplied energy changes the phase. A pure substance at fixed pressure can change phase without a temperature rise. Mixtures need not show the same constant-temperature plateau.

Use Q = mL for the amount changing phase. Here, m is mass in kg and L is specific latent heat in J/kg. Melting absorbs heat; freezing releases it.

Near one atmosphere, useful approximate values for water are:

Divide a value in J/kg by 1000 to express it in J/g. Values in calories are approximate and depend on the calorie definition.

Vaporisation requires overcoming intermolecular attractions. Water molecules do not split into hydrogen and oxygen atoms during ordinary boiling.

Follow the material's actual path

At ordinary pressure, ice below 0°C must first warm to its melting point. It then melts. The resulting water warms further only if the target temperature exceeds 0°C.

Warm ice → melt ice → warm liquid water, if needed.

Use mcΔT for each warming stage and mL for melting. Add the required stages. Latent heat is not automatically the first step.

Conduction, convection and radiation describe different transfer mechanisms. More than one mechanism can operate in the same situation.

• Conduction: energy transfers through microscopic interactions within matter.
• Convection: energy travels with the bulk movement of a fluid.
• Radiation: electromagnetic waves carry energy without requiring matter.

Trace a concrete heat-transfer path

Imagine a metal pot containing water on a heater. Energy passes through the metal base by conduction. Warmer water can rise while cooler water moves downward. This circulation distributes energy by convection.

The hot pot also emits thermal radiation toward its surroundings. Sunlight reaches Earth by radiation across space. Radiation does not require moving air or direct contact.

Conduction does not require the entire metal pot to move. Convection does require bulk fluid motion. Sublimation is a phase change, rather than a fourth heat-transfer mechanism.

Evaporation can cool the liquid left behind. Higher-energy molecules preferentially escape from the surface. Their escape can lower the remaining liquid's average molecular energy.

Sweat can cool skin as water evaporates. Water evaporating through an earthen pot's pores can also produce cooling. External heating can offset evaporative cooling.

Greater surface area, higher temperature, lower humidity and moving air generally encourage evaporation. Moving air removes moist air near the surface. High humidity reduces that opportunity for further evaporation.

Why does air often move inland during the day?

Unequal land and water heating supports a typical daytime sea breeze. Land warms faster than the nearby sea. Air above land warms and rises. Cooler air then moves inland from the sea.

A land breeze can occur at night as land cools faster. Other weather conditions can disrupt this local circulation.

Dark surfaces commonly absorb more sunlight than comparable light surfaces. That visible-light comparison does not determine absorption at every thermal-radiation wavelength.

Sublimation is direct change from solid to gas. Deposition is the reverse change. Dry ice is solid carbon dioxide. At normal atmospheric pressure, dry ice sublimes rather than forming liquid carbon dioxide.

Matter also has states beyond solids, liquids and gases. Plasma contains charged particles and occurs in stars. A Bose–Einstein condensate is a quantum state formed at extremely low temperatures. Satyendra Nath Bose's work helped establish its theoretical foundations.

Thermometers measure temperature, with ranges chosen for their purpose. Traditional clinical instruments may cover about 35°C to 42°C. Some laboratory instruments cover about −10°C to 110°C. These are instrument examples, rather than universal ranges.

About 37°C is a familiar average body temperature. It is not an exact reading for every healthy person.

Mechanical work can increase a body's internal energy. Joule's nineteenth-century paddle-wheel experiments connected mechanical work with warming. Falling weights turned paddles in water. The water warmed as the paddles did work on it.

An adiabatic process has no heat transfer: Q = 0. Insulation can help approximate this condition. Adiabatic does not mean that temperature must remain constant.

The second law explains why spontaneous net heat transfer runs from hot to cold. A refrigerator moves energy from its cold interior toward a warmer room. Its compressor needs external work to achieve that transfer.

The National Defence Academy and Naval Academy examination is abbreviated NDA & NA. Its General Ability Test is abbreviated GAT.

What if specific heat changes with temperature?

Suppose c(T) = C0 + αT during sensible heating from T1 to T2. Each small temperature interval uses its own specific heat. Therefore, add the small heat contributions by integration.

Q = m∫[T1 to T2] c(T)dT

Q = m(T2 − T1)[C0 + (α/2)(T1 + T2)]

Here, m is mass in kg. Use temperature values in the scale specified by the given function. With temperature in K, C0 has units J/(kg·K). The coefficient α has units J/(kg·K²).

C0 is a coefficient in the specific-heat function. It is different from the apparatus heat capacity C_cal. Assume no phase change or other energy input in this heating model.

NDA & NA (I) 2026 GAT: Q53.

How much warm water melts cold ice?

Take 5 g of ice at −20°C and water initially at 30°C. The target is fully liquid water at 0°C, at ordinary pressure. Neglect apparatus absorption and outside heat exchange in this example.

Use the problem's rounded values: c_ice = 0.5 cal/(g·°C), L_f = 80 cal/g, and c_water = 1 cal/(g·°C).

1. Warm the ice: Q_warm = 5 × 0.5 × 20 = 50 cal.
2. Melt the ice: Q_melt = 5 × 80 = 400 cal.
3. Add the gains: Q_ice = 50 + 400 = 450 cal.
4. Each gram of warm water releases 1 × 30 = 30 cal when cooling to 0°C.
5. Required water mass: 450/30 = 15 g = 0.015 kg.

The melted ice needs no further warming because the target is 0°C. Skipping the first stage underestimates the required water.

NDA & NA (I) 2026 GAT: Q61.