Physics › Thermal physics › Specific heat capacity and latent heat

Specific heat capacity and latent heat

You will learn how much energy it takes to warm a material, to melt it or to boil it, and how to measure that energy. The specific heat capacity is the energy that raises 1 kg of a material by 1 K, and Q = mcΔθ gives the energy for a temperature change. Specific latent heats of fusion and vaporisation apply during a change of state, through Q = ml. Both constants are measured electrically, with a second run used to cancel the heat losses.

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Year 13AQA 3.6.2.1 CIE 14.3.1, 14.3.2 OCR A 5.1.3(a), 5.1.3(b)(i), 5.1.3(c), 5.1.3(d)(i)EDEXCEL 2, 144

Thermal energy transfer and specific heat capacity, part 2 of 2. Part 1 is Internal energy and changes of state.

Builds on Internal energy and changes of state and Series, parallel and power.

Related elsewhere: measuring an enthalpy change: calorimetry on InkChemistry.

IN THIS TOPIC

Calculate energy transfers with Q = mcΔθ and Q = ml, stage by stage where a problem needs it.
Describe an electrical determination of a specific latent heat, for melting and for boiling, and say how the heat losses are cancelled.

COMMON MISCONCEPTION

Misconception:

A kilogram of steam at 100 °C and a kilogram of water at 100 °C carry the same energy.

Steam at 100 °C carries far more energy than water at the same temperature: turning the water into steam took a further 2260 kJ per kilogram, the specific latent heat of vaporisation, and that is why a steam scald is so much worse than a hot-water one.

Calculating a temperature rise

The specific heat capacity c of a material is the energy needed to raise the temperature of 1 kg of it by 1 K.

Q = mc\Delta\theta

IN THE AQA DATA BOOKLET

Two equal-mass columns receiving the same energy: water's temperature bar rises a little, aluminium's several times further, in exact inverse ratio to their specific heat capacities. — FIG. 1Equal masses, equal energy in: water barely warms while aluminium leaps, because c sets a different temperature rise for each material.

Water's enormous c, about 4200 J kg⁻¹ K⁻¹, lets the sea moderate coastal climates and makes a kettle take as long as it does. One practical method deserves knowing. In continuous-flow heating, liquid streams steadily past an electrical heater. Three things are measured: the heater power P, the mass m carried past each second (a mass flow rate, in kg s⁻¹), and the temperature rise Δθ from inlet to outlet. Each second the heater supplies P, the liquid takes mcΔθ of it, and the rest leaks to the surroundings as a loss h. One run therefore has two unknowns in it, c and h, so one run can never settle c.

The second run is what does. Change the flow rate and adjust the power until the inlet and outlet temperatures read exactly as before. The apparatus now sits at the same temperatures as in the first run, so the loss h is the same number, and subtracting the two power equations deletes it,

P_{1} − P_{2} = (m_{1} − m_{2})c\Delta\theta

with m₁ and m₂ the masses carried past per second in the two runs, flow rates in kg s⁻¹, so that both sides of the equation are powers, in watts. Divide the power difference by Δθ and by the difference in those two masses and the result is c, with the loss removed. The method needs both powers; the flow readings on their own carry no energy in them at all.

WORKED EXAMPLE

Sizing a shower

A 9.0 kW electric shower warms water by 25 K as it flows through. Find the flow rate it can sustain. (c = 4200 J kg⁻¹ K⁻¹.)

Power is energy per second, so apply Q = mcΔθ to one second of flow. That gives 9000 = m × 4200 × 25.

So m = 9000/105 000 = 0.086 kg each second, about five litres a minute.

Your own bathroom is the sense check. Electric showers run gentler than mains-pressure taps, and this is the arithmetic reason. More flow would mean less warming, because the wire fixes the energy budget.

Measuring specific heat capacity electrically

OCR A (H556) gives this as the source example for PAG11: determine the specific heat capacity of a material. It is referenced to 5.1.3(b)(i), an electrical experiment to determine the specific heat capacity of a metal or a liquid. Statement 5.1.3(b)(ii) adds techniques and procedures used for an electrical method to determine the specific heat capacity of a metal block and a liquid. CIE 9702 sets no list of practicals, but its Paper 3 and Paper 5 questions may be based on any physics, heating included. AQA examines specific heat capacity in 3.6.2.1.

Apparatus	Purpose
Aluminium block, about 1 kg, with two holes	Holds the heater and thermometer inside the metal
12 V immersion heater, about 50 W	A steady, known power
Thermometer, with oil in its hole	Oil fills the air gap for good thermal contact
Insulating jacket and heat-proof mat	Less energy lost
Joulemeter, or voltmeter and ammeter	Energy supplied, directly or as VIt
Balance and stopclock	Mass and time

Weigh the block, fit the heater and thermometer, and insulate it. Switch on and start the clock together, keep V and I steady, and record θ every 30 s for about ten minutes.

Plot θ against t. After a slow start the graph becomes a straight line. Each second the heater supplies VI joules, so on the straight part VI = mc × gradient, and

c = \frac{VI}{m \times \text{gradient}}

With a joulemeter, plot θ against energy supplied; the gradient is 1/mc.

A single rise from first reading to last is less reliable, for two reasons. At first the heater warms itself and the metal next to it before the thermometer responds, and this thermal lag shrinks a start-to-finish Δθ. Heat loss grows as the block warms above the room, and a gradient taken early in the run, near room temperature, keeps it small. Both errors make c come out too large.

Insulation reduces the loss. So does starting a few kelvin below room temperature and stopping the same amount above it: what the block gains from the room early on roughly balances what it loses later.

For a liquid, heat a known mass in a polystyrene cup, which takes too little energy to matter, and stir throughout. In a metal calorimeter, write VI = (mc + m_calc_cal) × gradient.

WORKED EXAMPLE

An aluminium block, from the gradient and from two readings

Idealised data. A heater at 12.0 V and 4.00 A warms a 1.000 kg aluminium block. Find c from the gradient, then from the end readings alone.

t / s	θ / °C
0	18.0
120	20.5
240	26.4
360	32.4
480	38.4
600	44.4

The power is VI = 48.0 W. From 240 s onwards θ rises 6.0 K in every 120 s, a gradient of 0.0500 K s⁻¹, so c = 48.0/(1.000 × 0.0500) = 960 J kg⁻¹ K⁻¹.

The end readings give c = (48.0 × 600)/(1.000 × 26.4) = 1.09 × 10³ J kg⁻¹ K⁻¹, 14% higher, almost all of it from the first two minutes, when θ rose 2.5 K against 6.0 K in each later interval.

The data-book value for aluminium is about 900 J kg⁻¹ K⁻¹. Heat loss and the heater's own warming would account for the gradient value sitting 7% above it.

Safety. The heater and block stay hot long after switching off. An immersion heater must not be switched on in air, where it can overheat and fail, so fit it into the block or liquid first. Follow your school's risk assessment.

Latent heat and changes of state

The specific latent heat l is the energy to change the state of 1 kg with no temperature change.

Q = ml

IN THE AQA DATA BOOKLET

Latent heat of fusion for melting, of vaporisation for boiling, and the second is far larger, since boiling must separate the particles almost completely. Use Q = mcΔθ when the temperature changes without a change of state, and Q = ml during a change of state at constant temperature. For a multi-stage process, warm the ice, melt it, warm the water, calculate the energy for each stage separately and add the results.

Both latent heats are measured electrically, and both measurements are built the same way. An immersion heater supplies a known power, P = VI, read from a voltmeter across the heater and an ammeter in series with it. The energy delivered in a timed run is then VIt. A balance weighs the substance that changed state. A good version of the experiment differs from a poor one in how it handles the energy the room adds or takes away. In both cases the answer is a second run, not more lagging.

For fusion, the heater sits in crushed melting ice held in a funnel, and the meltwater it produces is collected in a beaker on the balance for a measured time. Warm air melts the ice as well, so an identical funnel of ice stands beside the first with its heater switched off, draining into its own beaker over the same time. That control run measures the mass the room melts unaided, m₀, and the heater is then answerable only for the difference, so VIt = (m − m₀)l_f. The ice must be melting to begin with, since any energy spent warming it from a freezer temperature belongs to c and not to l.

For vaporisation, the liquid stands on a top-pan balance with the heater immersed in it. Readings begin only once it is boiling steadily. By then the apparatus has stopped warming up, so every extra joule leaves as vapour or as loss. Record the mass boiled away in a measured time, then repeat at a different heater power while the liquid boils just as steadily. The apparatus sits at the same temperature in both runs, so the loss to the room is the same in each. Subtracting the two equations removes it, just as in the continuous-flow method.

That leaves $l_{v} = (P_{1} - P_{2})t/(m_{1} - m_{2})$ , with m₁ and m₂ the masses boiled away in the same time t at the two heater powers, and the heat loss gone from the answer without ever having been measured.

WORKED EXAMPLE

A latent heat from two runs, and the loss it cancels

A heater in a boiling liquid runs at 60 W and the balance shows 12 g lost in 100 s. Raised to 100 W, with the liquid boiling just as steadily, it loses 24 g in the same 100 s. Find the specific latent heat of vaporisation, then the power going to the surroundings.

Subtract the runs. l_v = (100 − 60) × 100/(0.024 − 0.012) = 3.3 × 10⁵ J kg⁻¹, with the masses in kilograms.

Now go back to either run for the loss. In the first, boiling off 0.012 kg in 100 s costs (0.012 × 3.3 × 10⁵)/100 = 40 W, so the other 20 W of the 60 W went to the room.

The second run agrees, which is the check worth doing: 0.024 kg in 100 s costs 80 W, leaving the same 20 W. A third of the smaller heater's power never reached the liquid at all.

Two boiling runs at different heater powers: each run's power splits into the part carried off as vapour and the same loss to the room, so subtracting the runs leaves the latent heat with the loss gone — FIG. 2The same two runs drawn as power bars. Each heater's power splits into the part carried off as vapour, m l / t, and the part lost to the room, and the loss comes out as the same 20 W in both runs because the liquid is boiling just as steadily in each. Subtract the runs and that segment cancels: 40 W of extra power boiling off an extra 12 g in the same 100 s, which is the specific latent heat and nothing else.

GUIDED PRACTICE

Ice to drinking water

Find the energy to turn 0.50 kg of ice at 0 °C into water at 20 °C. (l_f = 3.3 × 10⁵ J kg⁻¹, c = 4200 J kg⁻¹ K⁻¹.) Work the two stages separately.

Show the working

Melting comes first, at constant temperature. Q = ml = 0.50 × 3.3 × 10⁵ = 1.65 × 10⁵ J.

Then warming the meltwater. Q = mcΔθ = 0.50 × 4200 × 20 = 4.2 × 10⁴ J.

Total 2.1 × 10⁵ J, with the melt costing roughly four times the warming. Latent heats are large compared with mcΔθ over a few tens of kelvin, which is why the state change dominates the total.

INDEPENDENT PRACTICE

Why sweating works

During exercise, 0.030 kg of sweat evaporates from a runner's skin. Find the energy carried away. (l_v for water at skin temperature is about 2.3 × 10⁶ J kg⁻¹.)

Show the working

Q = ml_v = 0.030 × 2.3 × 10⁶ = 6.9 × 10⁴ J.

Seventy kilojoules out of thirty grams. Evaporation is the body's most effective cooling channel, because latent heat dwarfs specific heat.

One kilogram of ice at zero degrees taken to steam at a hundred: boiling costs more than melting it and heating it through the whole hundred degrees put together — FIG. 3One kilogram of ice at 0 °C taken all the way to steam at 100 °C, and where the energy goes. Melting it costs 334 kJ, heating the water through the whole hundred degrees costs 420 kJ, and boiling it costs 2260 kJ, more than the other two put together and nearly seven times the melting. The plateau at the top of a heating curve is long for a reason.

ASSESSMENT FOCUS

Δθ is a temperature difference and comes out identical in °C and K, so no conversion is needed.
Multi-stage heating means one Q per stage, then a sum. Applying mcΔθ across a plateau, where ml belongs, charges the energy to a temperature change that does not happen.
In the electrical latent-heat experiments, the second run is the essential part. For melting, an identical funnel of ice with the heater off measures what the room melts, and that mass is subtracted. For boiling, a second power at the same steady boil holds the loss fixed so that subtracting the two runs removes it. A description of one run, with the discrepancy attributed to heat losses, does not account for them; only the second run removes them from the result.
Summarise the continuous-flow method in one line. Two runs, two flow rates, two powers, the same temperature rise, then subtract to eliminate the heat loss: P₁ − P₂ = (m₁ − m₂)cΔθ. The flow rates alone give no route to c, since c comes out of the power difference.

CHECK YOURSELF

A 2.2 kW kettle holds 0.80 kg of water at 15 °C. How long does it take to reach 100 °C? (c of water = 4200 J kg⁻¹ K⁻¹; assume no losses.)

Show a hint

Find the energy for the temperature rise first, then divide by the power.

Show the answer

Energy needed is Q = mcΔθ = 0.80 × 4200 × 85 = 2.86 × 10⁵ J.

Time follows from t = Q/P = 2.86 × 10⁵ / 2200 = 130 s, a little over two minutes.

A real kettle takes slightly longer, since some energy heats the kettle body and the room. That is the loss the continuous-flow method is built to cancel.

Q = mcΔθ while the temperature moves, Q = ml while it stands still, and a problem that crosses a state change is one Q per stage added up.

Δθ is a difference, so it is the same number in °C and in K and needs no conversion.

The electrical experiments cancel their losses with a second run: same temperature, different power, and the difference is the energy the substance took.

17 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

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Calculate energy transfers with Q = mcΔθ and Q = ml, stage by stage where a problem needs it.
Describe an electrical determination of a specific latent heat, for melting and for boiling, and say how the heat losses are cancelled.

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