Heat and thermal energy balance

Two identical burners heat a cup and a pan of water. The cup warms faster. How can we relate the energy received, water mass and temperature change in advance, and check where the energy went?

Place a cold spoon in hot water. The water loses energy and the spoon gains it as their temperatures approach each other. Energy transferred by heating is called heat Q and is measured in joules. It describes a process: the spoon's internal energy changes, but no separate substance called ‘heat’ is stored inside. For calculations, we take energy received as positive Q and energy given out as negative Q.

One kilogram of water needs a certain energy for a given temperature rise. Two such kilograms need twice as much, because both must be warmed. But a kilogram of another substance needs a different amount. Specific heat capacity c describes this difference: the energy needed to warm 1 kg of a substance by 1 K without a change of state. Its unit is J/(kg·K); mass m is measured in kg. A larger c means a smaller temperature rise for the same energy received.

We need the change in thermometer reading, not the reading itself: ΔT = Tкон − Tнач. During warming, the final value exceeds the initial value, so ΔT is positive; during cooling it is negative. Readings in °C can also be subtracted: the numerical difference is the same as in K. If initial and final temperatures coincide, ΔT = 0, but zero heat transfer follows only when there is no change of state.

Combine the observations: energy increases with mass m, the required temperature rise ΔT, and specific heat capacity c. This gives Q = cmΔT for nearly constant c and no phase change. Doubling m or ΔT doubles Q; halving the temperature rise requires half the energy. At ΔT = 0 the model gives Q = 0. Units agree too: Дж/(кг·К) · кг · К = Дж. During cooling, negative Q means energy is transferred out.

Mix hot and cold water. To find the final temperature, we need not track every molecule; accounting for the energy of both portions is enough. A calorimeter reduces exchange with the surroundings, bringing the experiment closer to an isolated system. With no external heat transfer or external work, all energy given out by the hot portion goes to the cold portion. This is energy conservation applied to heat exchange, not a new independent energy source.

Let T be the shared final temperature. For the cold object, T − T₁ is positive; for the hot object, T − T₂ is negative. The transfers sum to zero: Q₁ + Q₂ = 0, or c₁m₁(T − T₁) + c₂m₂(T − T₂) = 0. Keep the same sign convention throughout. Increasing one object's mass makes its temperature change less, bringing final T closer to its initial temperature. With equal initial temperatures there is no heat exchange.

What if the cold container is massive? It warms too, so some energy from the hot water goes into the walls rather than the other water portion. Include the container and thermometer as separate terms in the balance, or explicitly assume their heat capacities are negligible. With no phase changes or external sources, the final temperature lies between the initial temperatures of all participating objects. A result hotter than the hottest portion signals an error or an unaccounted source.

Where does a burner's energy come from? Combustion rearranges chemical bonds in fuel and oxygen; the reaction products have a different chemical energy. Specific energy of combustion q is the energy released by completely burning 1 kg of a fuel. Its unit is J/kg. It is a fuel property under specified conditions, not a property of the water above the flame. A larger q means that less fuel mass is needed to release the same energy.

If one kilogram releases q joules, mass m releases Q = qm in complete combustion. Double the mass and the energy doubles; at m = 0 there is no energy source from fuel. But not all of it reaches the water: the container and air warm up, and combustion products carry energy away. If the useful fraction, or efficiency, is η, the water receives ηqm. Use η as a fraction of one, not a percentage number. Incomplete combustion makes qm an overestimate of the energy released.

Mix 1 kg of water at 60 °C with 1 kg at 20 °C. Neglect losses and the container's heat capacity. Find the common temperature.

Find the temperature at which energy given out by the hot water equals energy received by the cold water. The masses and specific heat capacities are equal; the problem excludes external exchange and container heating.

Write the balance using one sign convention: c·1·(T − 60) + c·1·(T − 20) = 0. The first term will be negative and the second positive.

Cancel c to obtain 2T − 80 = 0 and T = 40 °C. The hot portion cools by 20 degrees and the cold one warms by 20. With equal masses, the energy transfers are equal.

The common temperature is 40 °C. It is midway between 20 and 60 °C precisely because the two portions have equal heat capacities.

Q describes energy transfer; c, m and ΔT determine its size during warming without a change of state.

An energy balance must include all participants and keep one consistent sign convention.

Complete combustion releases qm; only part of that energy may reach the object we want to heat.

During melting, energy enters while temperature may stay constant. The next lesson explains where that energy goes.