Latent heat and energy transfer
Changes of state and latent heat with Q = mL, reading a heating curve quantitatively, conduction and the rate equation Q/t = kAΔT/d, convection, radiation, and applying all three to insulation and cooling.
A burn from steam at 100 °C is far worse than a splash of boiling water at the same temperature. The reason is hidden in the moment the steam touches your skin and condenses: every gram releases about 2260 J, more than five times the energy that gram of water then gives up as it cools from 100 °C all the way to body temperature. That hidden energy is latent heat, and it also explains why sweating cools you down.
Latent heat
When a substance melts or boils, energy goes in but the temperature stays the same until the change of state is complete. The energy is not raising the average kinetic energy of the particles; it is being used to separate them, overcoming the attractions between them and increasing their potential energy. This energy is called latent heat ("latent" means hidden).
$Q = mL$
- The specific latent heat of fusion $L_f$ is the energy needed to melt 1 kg of a solid at its melting point. For water, $L_f = 3.34 \times 10^5$ J kg$^{-1}$.
- The specific latent heat of vaporisation $L_v$ is the energy needed to turn 1 kg of liquid into gas at its boiling point. For water, $L_v = 2.26 \times 10^6$ J kg$^{-1}$.
The same energy is released when the change goes the other way (freezing, condensing). $L_v$ is much bigger than $L_f$ because boiling separates the particles completely, while melting only loosens them.
Evaporation also needs latent heat. The fastest molecules escape from the surface, so the average kinetic energy of those left behind falls and the liquid cools. Sweat, the "Coolgardie safe" used in the Australian goldfields to keep food cool with wet hessian, and evaporative air coolers all use this.
Heating curves
On a heating curve with a steady energy supply (constant power $P$, so $Q = Pt$):
- On sloping sections one state is warming. The gradient is $\dfrac{\Delta T}{Q} = \dfrac{1}{mc}$, so a gentler slope means a larger specific heat capacity (water's slope is gentler than ice's).
- On flat sections the state is changing at constant temperature, and the length of the flat section is $mL$. The boiling plateau is about 6.8 times as long as the melting plateau for water.
How much energy is needed to turn 0.50 kg of ice at −20 °C into steam at 100 °C? ($c_{\text{ice}} = 2100$, $c_{\text{water}} = 4180$ J kg$^{-1}$ K$^{-1}$.)
Ice at 0 °C is added to 0.300 kg of soft drink (treat as water) at 25.0 °C. What mass of ice is needed to bring the drink to 5.0 °C? Ignore losses.
Conduction
Conduction is the transfer of energy through a material without the material itself moving. Faster-vibrating particles at the hot end collide with their neighbours and pass energy along. In metals, free (delocalised) electrons also carry energy quickly through the lattice, which is why metals are good conductors.
The rate of conduction through a slab is
$\dfrac{Q}{t} = \dfrac{kA\Delta T}{d}$
where $Q/t$ is the rate of energy transfer (W), $k$ the thermal conductivity of the material (W m$^{-1}$ K$^{-1}$), $A$ the cross-sectional area (m²), $\Delta T$ the temperature difference across the slab (K), and $d$ its thickness (m).
| Material | $k$ (W m$^{-1}$ K$^{-1}$) |
|---|---|
| Copper | 400 |
| Aluminium | about 240 (alloys lower) |
| Steel | about 50 |
| Glass | 0.8 |
| Brick | 0.6 |
| Wood | 0.12 |
| Fibreglass insulation | 0.04 |
| Still air | 0.025 |
A glass window is 1.5 m by 1.2 m and 4.0 mm thick. Inside the glass surface is at 18 °C and outside at 8 °C. Find the rate of energy loss by conduction.
In 10 minutes, 1.8 × 10⁴ J is conducted through a metal rod of cross-section 2.0 × 10⁻⁴ m² and length 0.25 m, whose ends are held at 100 °C and 20 °C. Find $k$.
Convection and radiation
Convection is energy transfer by the bulk movement of a fluid (liquid or gas). Fluid near a heat source warms, expands and becomes less dense, so it rises, and cooler, denser fluid sinks to take its place, forming a convection current. Convection drives sea breezes (land heats faster than sea, warm air rises over the land and cool sea air flows in), heats a room from a single heater, and circulates water in hot-water systems. Convection cannot happen in a solid, and it does not happen in a vacuum. Insulation such as fibreglass and double glazing works by trapping air in small spaces where it cannot circulate, leaving only slow conduction through still air.
Radiation is energy transfer by electromagnetic waves, mostly infrared at everyday temperatures. It needs no medium, which is how the Sun's energy reaches Earth. Every object emits radiation, and the hotter it is the more it emits (the rate rises very steeply, with the fourth power of the absolute temperature). Dark, matt surfaces are good absorbers and good emitters; shiny, light surfaces are poor at both. That is why many Australian roofs are now light-coloured ("cool roofs"), why emergency blankets are shiny, and why car radiators are painted black.
| Conduction | Convection | Radiation | |
|---|---|---|---|
| How | particle collisions, free electrons | bulk movement of fluid | EM (infrared) waves |
| Where | mainly solids | liquids and gases | anywhere, including a vacuum |
| Reduced by | low-$k$ materials, thickness | trapping fluid in small spaces | shiny surfaces |
A vacuum flask attacks all three: the vacuum between two glass walls stops conduction and convection, and silvered walls reduce radiation.
- In Conduction mode with glass, A = 1.0 m², d = 4 mm and ΔT = 20 °C, the rate is 4,000 W. Change d to 8 mm: it halves. Double A: it doubles back.
- Choose brick with d = 100 mm (A = 1.0 m², ΔT = 20 °C): 120 W. Choose copper with d = 4 mm: 2,000,000 W. Explain the difference using $k$.
- Turn on Double glazing: the rate falls to 40.8 W, compared with 4,000 W for a single pane. Which layer does almost all the work?
- In Ice to steam with 0.5 kg of ice and a 500 W heater, press Start. Read the times of the flat sections: melting 334 s, boiling 2,260 s. Check each with $t = mL/P$.
Test yourself
The energy needed to melt 250 g of ice at 0 °C is
$Q = mL_f = 0.250 \times 3.34 \times 10^5 = 8.35 \times 10^4$ J.
A wall is replaced by one of the same material that is twice as thick and has half the area. For the same temperature difference, the rate of conduction is
$Q/t \propto A/d$: half the area and twice the thickness gives $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
On the heating curve of a pure substance heated at constant power, the flat section for boiling is longer than the flat section for melting because
Time on a plateau is $mL/P$, so the longer plateau has the larger $L$.
Double glazing reduces energy loss mainly because
Still air has $k = 0.025$, about 30 times smaller than glass. A narrow sealed gap keeps convection small.
Label the heating curve.
Which transfer is mainly responsible?
During a change of state the stays constant while energy is added. The energy needed to boil 1 kg of a liquid at its boiling point is the specific latent heat of . The rate of conduction is proportional to the and the temperature difference. Convection only happens in . Radiation can travel through a .
A 1200 W kettle contains 0.80 kg of water at 20 °C. It is switched on and accidentally left on with its automatic switch broken. (a) Calculate the time to reach 100 °C. (b) Calculate how much longer it would take to boil all the water away. (c) Sketch and label a graph of temperature against time for the whole process. (d) The base of the kettle's element has an area of 0.010 m² and its metal sheath is 1.0 mm thick with $k = 16$ W m⁻¹ K⁻¹. Calculate the temperature difference across the sheath while it delivers 1200 W, and explain why kettle elements burn out if switched on empty.
(a) $Q = mc\Delta T = 0.80 \times 4180 \times 80 = 2.68 \times 10^5$ J; $t = Q/P = 2.68 \times 10^5/1200 = 223$ s (about 3.7 minutes).
(b) $Q = mL_v = 0.80 \times 2.26 \times 10^6 = 1.81 \times 10^6$ J; $t = 1.81 \times 10^6/1200 = 1.51 \times 10^3$ s, about 25 minutes more.
(c) Temperature rises in a straight line from 20 °C at $t = 0$ to 100 °C at 223 s, then stays flat at 100 °C (labelled "boiling: Q = mLv") until about 1730 s, when the water is gone.
(d) $\Delta T = \dfrac{(Q/t)\,d}{kA} = \dfrac{1200 \times 1.0 \times 10^{-3}}{16 \times 0.010} = 7.5$ K. With water present, the water absorbs 1200 W while staying at or below 100 °C, and convection carries energy away from the element, so the element stays only a little hotter than the water. If the kettle is empty there is only air, which has a tiny conductivity and specific heat capacity and no latent heat to absorb. Energy builds up in the element faster than it can escape, its temperature rises rapidly, and the wire or sheath overheats and fails.
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