Heat & Specific Heat

In everyday speech a hot object has heat in it. In physics it does not. Heat is energy crossing the boundary of a system because the system and its surroundings sit at different temperatures; once across, it is internal energy of whatever received it and carries no memory of how it arrived. Joule's paddle-wheel experiment made the point by raising water's temperature with falling weights alone, no temperature difference anywhere. That is why the unit of heat is the joule, and why the calorie, the energy that lifts one gram of water by one kelvin, is 4.186 J rather than a separate quantity. Fix a sign convention and hold it: Q > 0 when energy enters the system you named. Most calorimetry errors are really a system boundary that moved halfway through the problem.

Supply energy to a body at a steady rate and, as long as it neither melts nor boils nor leaks much to the room, its temperature climbs in a straight line, so the energy delivered is proportional to the rise. Plot the energy supplied against the temperature rise and the slope of that graph is m c; divide by the mass and you have the specific heat capacity, since Q = m c ΔT gives c = Q/(m ΔT), in J kg⁻¹ K⁻¹. Measuring it is direct. Drive a 50.0 W heater in a 0.500 kg aluminium block for 120 s; the electrical work is Q = P t = 6.00 × 10³ J and the block rises ΔT = Q/(m c) = 6000/(0.500 × 900) = 13.3 K. Run the same experiment backwards, measuring ΔT and solving for c, and you have the standard electrical determination. The numbers themselves carry physics. Water's 4186 J kg⁻¹ K⁻¹ is enormous, more than four times air's and nearly eleven times copper's 385, which is why water is the coolant of choice, why coastal air is mild, and why a pool takes all summer to warm.

Dividing by mass is one choice; dividing by the number of particles is another, and for solids it is the revealing one. The molar heat capacity is Cₘ = M c, with M the molar mass in kg mol⁻¹, so Q = n Cₘ ΔT for n moles. Per kilogram, aluminium at 900 J kg⁻¹ K⁻¹ and lead at 128 differ by a factor of seven. Per mole they nearly coincide: 900 × 0.02698 = 24.3 and 128 × 0.2072 = 26.5 J mol⁻¹ K⁻¹, with copper's 385 × 0.0635 giving 24.4. This is the Dulong-Petit regularity, most simple crystalline solids sitting near 3R = 24.9 J mol⁻¹ K⁻¹ at room temperature, and it says the store of energy is the atom rather than the kilogram: aluminium's advantage per kilogram is really the lightness of its atoms. The rule fails for light, stiffly bound atoms, diamond managing only about 6 J mol⁻¹ K⁻¹, and for any solid cooled far below room temperature, where Cₘ collapses towards zero.

Seal two bodies in a container that exchanges nothing with the room, and the energy leaving one must arrive at the other: Q₁ + Q₂ = 0. Write each term as m c (T_f − Tₛₜₐᵣₜ) with the signed temperature difference, and the hot body supplies its own minus sign, so nothing has to be rearranged by hand. One unknown is left, and solving gives T_f = (m₁c₁T₁ + m₂c₂T₂)/(m₁c₁ + m₂c₂), a weighted mean of the two starting temperatures with weights m c. Three checks come free. T_f must lie between T₁ and T₂. It must sit nearer the body with the larger m c, because that body needs more joules per kelvin and so moves less. And the two temperature changes are locked in inverse ratio, |ΔT₁|/|ΔT₂| = m₂c₂/(m₁c₁). Because every weight scales the same way, the equation extends to three or more bodies unchanged.

A calorimeter is not a spectator. The cup, the stirrer and the immersed stem of the thermometer all end at T_f, and each takes m c (T_f − Tₛₜₐᵣₜ) joules to get there. Lump them into one heat capacity C_cal = Σ m c, in J K⁻¹, and add it to the cold side of the balance; equivalently quote the water equivalent, C_cal/c_water, the mass of water that would behave identically. The correction is not cosmetic. A 0.080 kg copper cup is 30.8 J K⁻¹ beside 0.200 kg of water at 837 J K⁻¹, which is 3.5% of the cold side's total capacity of 868 J K⁻¹, so a specific heat deduced without it lands 3.5% low, a bias no amount of repeating the run will expose. Measure C_cal once, by heating the empty calorimeter electrically or by mixing two known masses of water in it, and reuse the number.

Four limits, in the order they usually bite. First, phase change: at 0 °C ice absorbs energy while the thermometer stands still, so m c ΔT returns zero for a real transfer and a melting or boiling stage needs a term of its own. Second, heat leaks: a real vessel trades energy with the room, and 10 K above ambient the loss can run to several per cent of the supply. Either cancel it, by starting as far below room temperature as the run will finish above it, or measure it from the cooling rate and correct. Third, c is not quite constant: water's shifts by about 1% between 0 and 100 °C, a metal's falls steeply at low temperature, and across hundreds of kelvin the product becomes Q = m ∫ c(T) dT. Fourth, the constraint matters for gases: air takes 1005 J kg⁻¹ K⁻¹ at constant pressure but only 718 J kg⁻¹ K⁻¹ at constant volume, since an expanding gas also does work on its surroundings. For solids and liquids the two differ by only a few per cent, which is why one value of c is quoted.