Physics › Thermal physics › Specific heat capacity and latent heat

Specific heat capacity and latent heat

The specific heat capacity of a material is the energy that raises one kilogram of it by one kelvin, and Q = mcΔθ covers a temperature change. Specific latent heat of fusion and of vaporisation take over at a plateau, through Q = ml. Both constants are measured electrically, a second run cancelling the losses.

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

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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.

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

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 1 kg of it by 1 K.

Q = mc\Delta\theta

ON THE AQA DATA SHEET

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, is what lets the sea moderate coastal climates and what 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, and three things get 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 rather than plain masses, 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 c drops out, loss and all. Both powers are the point of the method. 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.

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

ON THE AQA DATA SHEET

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, read as P = VI from a voltmeter across the heater and an ammeter in series with it, so the energy delivered in a timed run is VIt. A balance then weighs the substance that changed state. What separates a good version of the experiment from a poor one is entirely what it does about the energy the room is quietly adding or taking away, and in both cases the answer is a second run rather than 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 what the room does on its own, 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, and readings begin only once it is boiling steadily, so that the apparatus has stopped warming up and 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 number in each, and subtracting the two equations deletes it exactly as the continuous-flow method did.

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, and no single run could have told you that.

Three horizontal bars to one power scale. The first run's 60 W splits into 40 W carried off as vapour and 20 W lost to the room; the second run's 100 W splits into 80 W as vapour and the same 20 W to the room. The third bar is the second run minus the first, a bare 40 W with no loss segment left in it at all, because the loss was the same number in both runs and cancelled. That is why the experiment is done twice: the latent heat comes out of the difference, and the heat going to the room never has to be measured. — 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 powerful cooling channel, because latent heat dwarfs specific heat. That energy left with the vapour, taken from the skin that supplied it.

Three horizontal bars, drawn to one scale, for the energy each stage of taking a kilogram of ice at zero degrees to steam at a hundred degrees costs. Melting the ice takes 334 kilojoules and warming the water through the whole hundred degrees takes 420, but boiling it takes 2260, more than the other two together and nearly seven times the melting. The three come to 3014 kilojoules. — 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; the second run is what 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 exactly 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 rather than by lagging: same temperature, different power, and the difference is what the substance took.

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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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