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Conduction, convection and radiation, specific heat and calorimetry, and rates of heat flow through windows, walls and skin.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute heat flow by conduction and radiation and find temperatures after heat is exchanged.
From the earlier lessons you know that heat is energy flowing because of a temperature difference and that internal energy is tied to temperature. You know power is energy per time, in watts. This lesson asks how fast heat flows, and how much a given amount of heat changes a temperature.
| Term | What it means |
|---|---|
| Conduction | Heat passed through a material by collisions, $P = kA\Delta T/L$. |
| Thermal conductivity, k | How readily a material conducts heat, in W/(m·K). |
| Convection | Heat carried by the bulk motion of a fluid. |
| Radiation | Heat carried by electromagnetic waves, $P = e\sigma AT^4$. |
| Specific heat, c | Energy to raise one kilogram by one kelvin, in J/(kg·K). |
| Emissivity, e | How well a surface radiates, from $0$ to $1$. |
| R-value | Resistance to heat flow of a layer, $R = L/k$. |
Heat always flows from hot to cold, by three routes:
Heat $Q$ entering a mass $m$ raises its temperature by $\Delta T = Q/(mc)$, where $c$ is the specific heat. In an insulated mixture, heat lost by the hot parts equals heat gained by the cold ones.
Another way: picture
Picture standing near a campfire. The poker in your hand grows hot along its length: conduction. Smoke and sparks rise above the fire: convection. Your face feels warm even though the air between you and the flames is cool: radiation. One fire, three routes.
Another way: steps
In a solid, fast-vibrating atoms at the hot end jostle their neighbors, passing energy along. In metals, free electrons carry energy too, which is why metals conduct heat so well: copper's conductivity is about $400$ W/(m·K), glass's about $0.8$, and still air's only $0.026$.
The rate $P = kA\Delta T/L$ has a simple logic. Double the area and twice as many paths carry heat. Double the temperature difference and each path carries twice as much. Double the thickness and the heat must travel twice as far, halving the rate. Good insulators work by trapping air in small pockets where it cannot move.
In liquids and gases, heated fluid expands, becomes less dense and rises, while cooler fluid sinks to take its place. The resulting circulation carries heat far faster than conduction through still fluid. A room warms from a radiator because air circulates past it, not because heat conducts through the air.
Forced convection uses a fan or pump: a car's radiator, a computer's cooling fan, a convection oven. Wind speeds convection from your skin, which is why the National Weather Service reports a wind chill: on a $0$ °F day, a $15$ mph wind makes skin lose heat as fast as still air at $-19$ °F.
Every object above absolute zero emits electromagnetic radiation, at a rate that climbs as the fourth power of absolute temperature. At room temperature it is infrared, invisible to the eye but seen by thermal cameras. Hotter objects glow red, then white.
An object also absorbs radiation from its surroundings, so the net rate is $e\sigma A(T^4 - T_s^4)$. Radiation needs no medium: it is how the Sun heats the Earth across $150$ million km of vacuum. Shiny surfaces have low emissivity and radiate poorly, which is why thermos flasks are silvered and emergency blankets are foil.
The same heat warms different materials by different amounts. Water's specific heat, $4186$ J/(kg·K), is unusually large; iron's is about $450$, lead's $130$. A kilogram of water needs nearly ten times as much heat as a kilogram of iron to warm by one kelvin.
When a hot object is dropped into cool water in an insulated cup, energy is conserved: heat lost by the object equals heat gained by the water. Setting them equal gives the final temperature, a weighted average in which each part counts by its $mc$. Because water's $mc$ is large, the final temperature ends close to the water's.
Checking an answer. A final temperature must lie between the starting ones. Conduction must slow with thickness. Radiation rates must use kelvins, and a body warmer than its surroundings must lose net heat.
The conduction law is an experimental rule found by Joseph Fourier in 1822; it holds when the temperature changes steadily through the layer. Radiation's fourth-power law, found by Josef Stefan and explained by Ludwig Boltzmann, follows from treating light as a gas of photons in thermal equilibrium.
Calorimetry assumes an insulated system with no phase changes, so all heat goes into temperature. Specific heats vary a little with temperature, but over ordinary ranges treating them as constant is accurate to a few percent.
Touch a metal railing and a wooden bench on a cold morning. Both are at the air's temperature, yet the metal feels much colder. It conducts heat away from your hand far faster, so your skin cools more. Your fingers sense the rate of heat flow, not temperature.
The same effect makes tile floors feel colder than carpet and makes a metal spoon in hot soup too hot to hold while a wooden one stays cool. Given time, everything reaches the same temperature, but the rates on the way differ enormously.
Water's large specific heat moderates climates. The Pacific Ocean absorbs summer heat and releases it in winter, which is why San Francisco's temperatures vary far less through the year than Kansas City's, at a similar latitude.
Oceans have absorbed over ninety percent of the extra heat trapped by greenhouse gases since the 1970s. Their enormous heat capacity slows surface warming but commits the planet to further warming even if emissions stopped, because the ocean releases stored heat slowly for centuries.
American builders rate insulation by R-value, the resistance to conduction of a layer, in units of ft²·°F·h/BTU. One American R-unit equals $0.176$ m²·K/W. The Department of Energy recommends R-49 to R-60 for attics in cold northern states and R-30 in the South.
Because the heat loss is $A\Delta T/R$, doubling the R-value halves the loss. Insulation works by trapping air in fiberglass, cellulose or foam, and it is among the cheapest ways to cut heating bills. Windows are the weak point: even a good double-pane window is only about R-3, a tenth of an insulated wall.
Because every warm object radiates infrared in proportion to $T^4$, a camera sensitive to infrared sees temperature. Home energy auditors use them to find missing insulation, which shows as bright patches on a cold wall. Firefighters use them to find people through smoke.
Doctors and veterinarians use thermal imaging to spot inflammation, and power companies fly them over transmission lines to find overheating connections before they fail. The camera measures radiated power and converts it to temperature using the surface's emissivity, which is why shiny metal can fool it into reading cold.
Heat does not always raise a temperature. When ice melts or water boils, the heat goes into breaking the bonds that hold molecules in place, and the temperature stays fixed until the change is complete. The energy per kilogram is the latent heat: $334$ kJ/kg to melt ice and $2260$ kJ/kg to boil water, more than five times the energy needed to warm the same water from freezing to boiling.
That huge latent heat of vaporization is why sweating cools so effectively and why steam burns are worse than burns from boiling water: condensing steam on skin releases its latent heat before the water even begins to cool. In calorimetry with ice, the balance must include the heat $mL$ to melt it as well as the heat to warm the meltwater afterward.
In a cold-climate American house, much of the winter heat loss goes up through the attic floor. With $150$ m² of ceiling insulated to R-13, an old standard, and the living space $22$ K warmer than the attic, heat leaks at about $1400$ W, like leaving a space heater running around the clock.
Blowing in more cellulose to reach R-49, as the Department of Energy recommends for Minnesota, cuts that to under $400$ W, saving about $25$ kWh a day. Utilities in many states offer rebates for attic insulation because it is one of the cheapest ways to reduce energy use. Sealing air leaks around light fixtures and attic hatches matters too, since convection through gaps bypasses the insulation entirely.
An American electric kettle plugs into a $120$ V outlet and draws about $1500$ W, the most a standard household circuit comfortably supplies. Heating $1.5$ kg of water from $20$ °C to boiling needs $mc\Delta T = 502$ kJ, so it takes at least $335$ s, over five and a half minutes.
In Britain, where outlets supply $230$ V, kettles draw $3000$ W and boil the same water in half the time, which is one reason electric kettles are nearly universal there and less common in the United States. Real kettles take a little longer than the ideal time because some heat warms the kettle itself and escapes by convection and radiation from its sides.
It is natural to say that cold comes in through a window or that ice gives its cold to a drink. Physically, only heat flows, and always from hot to cold: the drink loses heat to the ice, and the room loses heat through the window. Cold is simply a lower temperature, with less internal energy.
A related error is to think objects that feel colder are colder. A metal railing and a wooden bench outdoors are at the same temperature; the metal feels colder because it conducts heat from your hand faster.
A single-pane window is $1.2$ m² of glass $4.0$ mm thick, with $k = 0.80$ W/(m·K). Convert the thickness.
$L = 0.0040\ \text{m}$
SI units.
Inside is $20$ °C, outside $-5$ °C. Find the difference.
$\Delta T = 25\ \text{K}$
Differences in °C and K agree.
Apply the conduction law.
$P = \dfrac{0.80 \times 1.2 \times 25}{0.0040} = 6000\ \text{W}$
An overestimate: air films slow it.
Replace it with double glazing of total R $0.35$ m²·K/W. Find the new loss.
$P = \dfrac{1.2 \times 25}{0.35} = 86\ \text{W}$
The trapped gas layer does the work.
Find the energy saved in a day.
$(6000 - 86) \times 24 \div 1000 = 142\ \text{kWh}$
An upper bound; real single panes lose less.
A $1.5$ kg iron skillet ($c = 450$ J/(kg·K)) at $200$ °C is plunged into $4.0$ kg of water at $20$ °C. Write the balance.
$1.5 \times 450 \times (200 - T) = 4.0 \times 4186 \times (T - 20)$
Heat lost equals heat gained.
Evaluate the heat capacities.
$675(200 - T) = 16{,}744(T - 20)$
The water's is far larger.
Expand both sides.
$135{,}000 - 675T = 16{,}744T - 334{,}880$
Distribute.
Collect the T terms.
$469{,}880 = 17{,}419T$
Move $T$ terms together.
Solve for the final temperature.
$T = 27.0\ \text{°C}$
Close to the water's start.
Find the heat transferred.
$Q = 16{,}744 \times 7.0 = 117\ \text{kJ}$
Gained by the water.
The Sun's surface is at $5800$ K with radius $6.96 \times 10^8$ m. Find its area.
$A = 4\pi r^2 = 6.09 \times 10^{18}\ \text{m}^2$
A sphere.
Raise the temperature to the fourth power.
$5800^4 = 1.13 \times 10^{15}$
In K⁴.
Apply the radiation law with $e = 1$.
$P = 5.67 \times 10^{-8} \times 6.09 \times 10^{18} \times 1.13 \times 10^{15}$
A near-perfect emitter.
Evaluate the power.
$P = 3.9 \times 10^{26}\ \text{W}$
The Sun's luminosity.
Spread it over a sphere at Earth's distance.
$\dfrac{3.9 \times 10^{26}}{4\pi (1.5 \times 10^{11})^2} = 1380\ \text{W/m}^2$
The solar constant.
Relate it to solar panels.
$\text{about } 1000\ \text{W/m}^2 \text{ reaches the ground}$
After passing through the atmosphere.
Write the heat formula.
$Q = mc\Delta T$
Specific heat of water.
Substitute the values.
$Q = 2.0 \times 4186 \times 15$
In joules.
Evaluate the heat.
A window is replaced by one with $1.5$ times the area and glass $2$ times as thick. With the same temperatures inside and out, by what factor does the rate of heat conduction change?
Complete the worked solution: a person's bare skin, area $2.0$ m² and emissivity $0.97$, is at $306$ K in a room whose walls are at $293$ K. With $\sigma = 5.67 \times 10^{-8}$ W/(m²·K⁴), find the power the skin radiates, the power it absorbs from the walls, and the net rate of loss, all in W.
Find the power radiated by the skin.
$P_{\text{out}} = 0.97 \times 5.67 \times 10^{-8} \times 2.0 \times 306^4 =$ e
Every surface radiates.
Find the power absorbed from the walls.
$P_{\text{in}} = 0.97 \times 5.67 \times 10^{-8} \times 2.0 \times 293^4 =$ b
The walls radiate too.
Subtract for the net loss.
$P_{\text{net}} = P_{\text{out}} - P_{\text{in}} =$ n
The difference matters.
Compare with metabolism.
$\text{about } 100\ \text{W at rest}$
Clothing cuts the loss.
Match each term to its description.
| energy passed by collisions through a material | energy carried by a moving fluid | energy carried by electromagnetic waves | energy to warm one kilogram by one kelvin | |
|---|---|---|---|---|
| conduction | ||||
| convection | ||||
| radiation | ||||
| specific heat |
A $0.50$ kg block of aluminum ($c = 900$ J/(kg·K)) at $100$ °C is dropped into $1.0$ kg of water ($c = 4186$ J/(kg·K)) at $20$ °C in an insulated cup. Fill in the final temperature in °C, the heat gained by the water in kJ, and the drop in the block's temperature in K.
| value | |
|---|---|
| final temperature (°C) | |
| heat gained by the water (kJ) | |
| drop in the block's temperature (K) |
Heat conducts through a glass window of area $1.5$ m² and thickness $0.004$ m, made of material with conductivity $0.80$ W/(m·K). Write the rate of heat flow, in W, as a function of the temperature difference $T$ across it, in kelvins.
Answer:
An electric kettle rated at $1200$ W heats $1.7$ kg of water from $20$ °C to $100$ °C. Assuming all the electrical energy goes into the water, how long does it take, in seconds? Water's specific heat is $4186$ J/(kg·K).
Answer: s
A house in Minneapolis has $150$ m² of attic floor insulated to R-$38$, which is $6.7$ m²·K/W in SI units. On a winter night the living space is $22$ K warmer than the attic. At what rate, in W, does heat flow up through the insulation?
Answer: W
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Heat conducts through a glass door of area $2.0$ m² and thickness $0.005$ m, made of material with conductivity $0.80$ W/(m·K). Write the rate of heat flow, in W, as a function of the temperature difference $T$ across it, in kelvins.
Answer:
You can analyze heat transfer. Explain to someone why a metal railing feels colder than a wooden bench at the same temperature.
22. Your turn: how much heat raises $2.0$ kg of water by $15$ K?, step 3
$Q = 125{,}580\ \text{J}$
About $126$ kJ.