Walk barefoot on Juhu beach at noon in May. The dry sand burns your soles, and you run for the wet strip near the water. Step into the sea and it feels cool. The same Sun has shone on the sand and on the sea all morning. Yet one is too hot to stand on and the other is pleasant.
Heat is energy on the move
Put a hot steel tumbler of tea into a bowl of cold water. The tea cools and the water warms until both show the same temperature. Something has passed from the tea to the water. That something is heat: energy that flows from one body to another because their temperatures differ. The chapter on temperature showed how a thermometer measures the temperature; here we measure the energy that flows.
A body does not contain heat, in the way it contains mass. It contains energy in the random motion of its molecules. Heat is the name of that energy only while it crosses from one body to another. So we say "the water gained 500 J of heat", never "the water has 500 J of heat".
Heat is the energy that flows from one body to another because of a difference in their temperatures. It always flows from the hotter body to the colder one, until their temperatures are equal.
Two units: the calorie and the joule
Heat was measured long before anyone knew it was energy. The old unit is the calorie: the heat needed to warm one gram of water by one degree Celsius. The SI unit is the joule, the unit of all energy. The two are joined by one number, found by experiment. This chapter uses \(1\,\text{cal} = 4.2\,\text{J}\); the measured value is 4.186 J.
The food Calorie on a biscuit packet, with a capital C, is a kilocalorie, 1000 calories. So a biscuit of 100 Calories can supply \(100\,000 \times 4.2 = 420\,000\,\text{J}\), or 420 kJ, to the body.
Joule's paddle wheel: work becomes heat
In the 1840s James Prescott Joule let two weights fall slowly through a measured height. Their strings turned a paddle wheel inside an insulated can of water. The churning paddles warmed the water by a fraction of a degree. The work done by the falling weights, \(mgh\), and the heat that appeared in the water always came in the same ratio. This fixed number is the mechanical equivalent of heat, \(J = W/Q = 4.2\,\text{J/cal}\).
So work and heat are two ways of handing energy to a body. A drill bit gets hot in a wall for the same reason as Joule's water. So do the brakes of a bus on a ghat road. The work done against friction appears as heat.
Water falls \(100\,\text{m}\) and comes to rest in the pool below. Suppose all its potential energy becomes heat in the water itself. Take \(g = 10\,\text{m/s}^2\) and \(c = 4200\,\text{J/(kg K)}\) for water. By how much does its temperature rise?
Take \(1\,\text{kg}\) of water. It loses \(mgh = 1 \times 10 \times 100 = 1000\,\text{J}\) of potential energy.
That energy warms the same kilogram: \(1000 = 1 \times 4200 \times \Delta T\), so \(\Delta T = 1000/4200 \approx 0.24\,\text{K}\).
The mass cancels, so every kilogram warms by the same 0.24 K, whatever the size of the fall. A tall fall is only very slightly warmer at the foot.
At Jog Falls in Karnataka the water drops about \(253\,\text{m}\). If all its potential energy became heat in the water, how much warmer would it be at the foot? Take \(g = 10\,\text{m/s}^2\) and \(c = 4200\,\text{J/(kg K)}\).
The heat to warm a body
How much heat warms a body? Twice the mass needs twice the heat for the same rise. Twice the rise needs twice the heat for the same mass. The third factor is the material. A kilogram of water needs about ten times the heat that a kilogram of copper needs for the same rise. So the heat is proportional to the mass \(m\) and to the rise \(\Delta T\), times a number for the material.
The specific heat capacity \(c\) of a material is the heat needed to raise the temperature of one kilogram of it by one kelvin: \(c = Q/(m\,\Delta T)\). Its SI unit is J/(kg K).
A change of one kelvin is the same size as a change of one degree Celsius. So \(\Delta T\) may be read on either scale; only the temperatures themselves differ by 273. In calories, water has \(c = 1\,\text{cal/(g °C)}\), by the very definition of the calorie. That is \(4.2\,\text{J}\) per \(10^{-3}\,\text{kg}\) per kelvin, or \(4200\,\text{J/(kg K)}\), the value this chapter uses.
Some specific heats, in J/(kg K), used in this chapter:
- water 4200, ice 2100, steam 2010;
- cooking oil about 2100, dry sand about 800;
- aluminium 900, iron 460, copper 400, mercury 140.
Water has one of the largest specific heats of any common substance. A kilogram of water takes 4200 J to warm by one kelvin; a kilogram of copper takes only 400 J.
Equal heat, unequal rise
A heater or a burner hands over heat at a steady rate, its power \(P\). In a time \(t\) it supplies \(Q = Pt\). If none is lost, \(Pt = mc\,\Delta T\). So the temperature rises at a steady rate, \(\Delta T / t = P/(mc)\). The smaller the specific heat, the faster the rise.
Read Fig 1.1 with the formula. Each liquid receives \(700 \times 120 = 84\,000\,\text{J}\). The water warms by \(84\,000/4200 = 20\,\text{K}\), and the oil by \(84\,000/2100 = 40\,\text{K}\). On the graph the oil's line is twice as steep, because its specific heat is half. For equal masses on equal heaters, the ratio of the slopes is the inverse ratio of the specific heats.
An electric kettle of \(1500\,\text{W}\) holds \(1\,\text{kg}\) of water at \(25\,\text{°C}\). Ignore heat lost to the kettle and the air. How long does it take to bring the water to \(100\,\text{°C}\)? Take \(c = 4200\,\text{J/(kg K)}\).
Heat needed: \(Q = mc\,\Delta T = 1 \times 4200 \times 75 = 315\,000\,\text{J}\).
Time: \(t = Q/P = 315\,000/1500 = 210\,\text{s}\), three and a half minutes.
A \(500\,\text{W}\) immersion rod heats a bucket of \(2\,\text{kg}\) of water for \(168\,\text{s}\). No heat is lost. By how much does the water warm? Take \(c = 4200\,\text{J/(kg K)}\).
Equal masses of two liquids, A and B, stand on identical heaters. A warms by \(40\,\text{K}\) in \(100\,\text{s}\) and B warms by \(20\,\text{K}\) in \(200\,\text{s}\). What is the ratio of their specific heats, \(c_A : c_B\)?
Why the sand burns and the sea stays cool
Now return to Juhu. Each square metre of sand and of sea receives the same sunlight. Dry sand has \(c \approx 800\,\text{J/(kg K)}\). The same heat that warms a kilogram of water by 10 K warms a kilogram of sand by \(4200/800 \times 10 \approx 52\,\text{K}\), more than five times as much. Two more effects help the sea. The Sun's heat stays in the top few centimetres of the sand, while waves and currents mix it through metres of water. And some of the sea's heat goes into evaporation.
The same large specific heat shapes the climate. The Arabian Sea beside Mumbai soaks up heat on summer days and gives it back on winter nights. So Mumbai's temperature changes little through the day and the year. Delhi, far from any sea, swings from about 45 °C in June to a few degrees in January.
Heat capacity: the whole body at once
Often we care about one particular body, not a kilogram of its material. The heat capacity \(C\) of a body is the heat needed to warm the whole body by one kelvin. For a body of one material, \(C = mc\), and then \(Q = C\,\Delta T\). A 200 g copper calorimeter has \(C = 0.2 \times 400 = 80\,\text{J/K}\). Specific heat belongs to a material; heat capacity belongs to an object.
A graph of \(Q\) against \(\Delta T\) is a straight line through the origin, as in Fig 1.2. Its slope is \(Q/\Delta T = C\). The steepest line belongs to the body that needs the most heat for each kelvin. For 1 kg, the slope is the specific heat itself. So from such a graph you can read \(C\), and dividing by the mass gives \(c\).
Per mole: the molar heat capacity
Chemists count substances in moles. The molar heat capacity is the heat needed to warm one mole by one kelvin. One mole has a mass \(M\), the molar mass, so the molar heat capacity is \(Mc\). For water, \(M = 18\,\text{g}\), so \(0.018 \times 4200 = 75.6\,\text{J/(mol K)}\). For aluminium, \(M = 27\,\text{g}\) and \(0.027 \times 900 = 24.3\,\text{J/(mol K)}\).
Work out the molar heat capacity of copper: \(0.0635 \times 400 = 25.4\,\text{J/(mol K)}\), close to aluminium's 24.3. Most solid metals come out near \(25\,\text{J/(mol K)}\). This is the law of Dulong and Petit (1819). The chapter on kinetic theory explains it: each atom of a solid shares energy equally among its ways of vibrating, which gives \(3R \approx 24.9\,\text{J/(mol K)}\) per mole.
The water equivalent
In a mixing experiment the vessel itself warms along with the water. It is simpler to count the vessel as extra water. The water equivalent \(W\) of a body is the mass of water that has the same heat capacity: \(W c_w = mc\), so \(W = mc/c_w\). The 200 g copper calorimeter has \(W = 80/4200 = 0.019\,\text{kg}\). It behaves as 19 g of water.
A \(200\,\text{g}\) copper calorimeter (\(c = 400\,\text{J/(kg K)}\)) holds \(100\,\text{g}\) of water. How much heat warms both by \(10\,\text{K}\)? Take \(c_w = 4200\,\text{J/(kg K)}\).
The calorimeter counts as \(80/4200\,\text{kg}\) of water, about 19 g. So heat \(119\,\text{g}\) of water: \((0.1 + 0.019) \times 4200 \times 10 \approx 5000\,\text{J}\).
Without rounding: \((420 + 80) \times 10 = 5000\,\text{J}\) exactly. The water takes 4200 J and the copper 800 J.
An aluminium pan has a mass of \(500\,\text{g}\) and \(c = 900\,\text{J/(kg K)}\). What is its water equivalent, in grams? Take \(c_w = 4200\,\text{J/(kg K)}\).
Rank these bodies by heat capacity, largest first.
Joule was the son of a Manchester brewer, and brewing taught him to measure temperatures to a hundredth of a degree. In 1843 he claimed that work and heat were the same thing, and most scientists ignored him. He spent the next seven years on the paddle-wheel experiment, refining it until his number was beyond doubt. The story goes that he carried a thermometer on his honeymoon in the Alps, to test whether a waterfall is warmer at the foot. The unit of energy carries his name.
- Saying a hot body "contains a lot of heat".A body contains energy. Heat is energy only while it flows because of a temperature difference.
- Converting a rise of 20 °C into 293 K before using \(Q = mc\,\Delta T\).A change of 1 °C is a change of 1 K. Only temperatures, not changes, differ by 273.
- Using grams with \(c\) in J/(kg K).Put the mass in kilograms, or use \(c\) in cal/(g °C) with grams throughout.
- Taking the body with the larger temperature rise to have the larger specific heat.For the same heat and mass, a larger \(c\) means a smaller rise.