### Episode 607: Specific heat capacity

Energy must be supplied (or rejected) to increase (or decrease) the temperature of a material. Here is how to calculate how much.

 SummaryDiscussion: Energy and change of phase. (15 minutes)Student experiment: Measuring specific heat capacities. (40 minutes)Worked example: Calculation involving c. (10 minutes)Student questions: Calculations. (30 minutes)

Discussion: Energy and change of phase
Up until this point the link between ‘internal energy’ and temperature has been qualitative, except for gases. In order to extend the discussion to solids and liquids we need to get more quantitative in two ways. One is to discuss how much the temperature of a body changes when its internal energy in increased by a certain amount. The other is to ask what happens when a substance changes phase from a solid to a liquid or liquid to a gas.

Start by introducing the equation for specific heat capacity c (SHC) and defining the terms. The word specific is an old fashioned way of saying ‘per unit mass’. Work through a simple calculation.

Understanding this equation will help solidify ideas about temperature and energy and how they differ. A possible analogy was supplied by Richard Feynman. He suggested thinking of heat energy as being like water, and temperature as wetness. A towel can have different amounts of fluffiness, so take more or less water to make it wet. When we dry ourselves, we dry until the towel is as wet as we are (“same temperature”).

The anomalously large SHC of water should also be discussed as it is particularly important for the development and maintenance of life on Earth.

NB nomenclature: there isn’t any agreed way to name c. Some use specific thermal capacity, others favour specific heating capacity to emphasise the fact that ‘heat’ is not an entity but a short hand name for a process (heating as oppose to working). Perhaps the most common is specific heat capacity.

Another source of confusion is treating state and phase as synonyms (as in changes of state / phase). Solids, liquids and gases are three of the different phases of matter (superfluids and plasmas are two others. NB Here, by a plasma, we mean an ionised gas, not a biological fluid). Thus melting, boiling etc are changes of phase. Each phase can exist in a variety of states depending upon e.g. the temperature and pressure. Thus the Ideal Gas Equation of State PV = nRT summarises the physically possible combinations of P, V and T for n moles of the ideal gas.

Student experiment: Measuring specific heat capacities
Students should carry out an experiment to measure the specific heat capacity of a solid and/or a liquid very soon after meeting the expression. There are a number of points to note here:

• If specific heat capacity is constant, the temperature will rise at a uniform rate so long as the power input is constant and no energy is lost to the outside
• There are large potential heat losses if the substance is not well insulated. These can be accounted for but in most cases students will not do so quantitatively
• They should calculate their value and make a comparison with data book values. They should be able to think of a number of reasons why their value does not match that in the data boo

Several different methods of determining the SHC of liquids and solids are given in the links below. Choose those best suited to your pupils and available equipment.

TAP 607-1: Measuring the specific heat capacity of a metal

TAP 607-2: The specific heat capacity of water and aluminium

It is useful to compare electrical methods of measuring the specific heat capacity of a solid and liquid including the continuous flow calorimeter for a liquid.

TAP 607-3: Measurement of specific heat capacities

Worked example:Calculation involving C
This example deals with the mixing of liquids at different temperatures.

• 5 l of water from a kettle at 90°C is mixed with a bucket of cold water (10 l at 10 °C) to warm it up for washing a car. Find the temperature of the mixed water, assuming no significant heat loss during the mixing. c of water = 4.2 kJ kg-1 °C -1

Answer: If there is no heat loss and no work done then the total energy of the system at the end is the same as at the start. Working with temperature differences from 10°C we have:

Initial energy = m c DT = 1.5 ´ 4200 ´ (90-10) [as 1.5 l has mass 1.5 kg] = 504 kJ

This is the amount of energy available to raise the temperature of all 11.5 l of water. Hence:

504 ´ 103 = 11.5 ´ 4200 ´ (T-10) (where T is the final temperature)

T = 20.4°C.

You can invent similar problems using a mixture of substances – a hot brick in water for example. However, be careful to consider realistic situations where not too much is lost as steam.

Student questions: Calculations
A range of questions involving specific heat capacity.

TAP 607-4: Specific heat capacity: some questions

TAP 607-5: Thermal changes