Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: heat, temperature and heat transfer (203-SN3-RE)

The heat transfer chapter asks for few formulas and a lot of rigour about what they stand for: a temperature or a difference, a flow or a resistance, a sensible or a latent heat. It is almost always there, and not in the calculation, that marks are lost.

This sheet gathers the traps found in 203-SN3-RE tests, the tree that picks the law from the form of the question, two model write-ups and a composite wall solved step by step. It is written for CEGEP students in the Natural Science program (Sciences de la nature).

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Temperature is a state, heat is a transfer: a body does not contain heat, it lets heat pass, from hot to cold, at a rate set by a resistance. Every trap in the chapter mixes up a state with a transfer, or a difference with a temperature.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

Temperature, thermal energy and heat

• TEMPERATURE is a state: it measures the average agitation of the particles. A difference of $1$ °C is a difference of $1$ K; an absolute temperature is $T=\theta+273.15$ .
• HEAT $Q$ is energy in transit, exchanged because two bodies are not at the same temperature. A body does not “contain” heat.
• Sensible heat: $Q=mc\,\Delta T$ , the temperature changes. Latent heat: $Q=mL$ , the state changes and the temperature stays constant.
• Calorimetry in an insulated enclosure: $\sum m_{i}c_{i}(T_{f}-T_{i})+\sum m_{j}L_{j}=0$ , every term with its sign.

To scale for

1

kg: the boiling plateau (

2260

kJ) is almost seven times longer than the melting plateau (

333

kJ); on the plateaus, the energy received does not change the temperature.

On a heating curve, the plateaus are the moments when the energy received does not change the temperature: it breaks bonds. A calculation that crosses a plateau without an mL term is wrong.

The three modes and their laws

• Conduction (in a solid or a still fluid): $\Phi=\dfrac{kA\,\Delta T}{e}=\dfrac{\Delta T}{R}$ , with $R=\dfrac{e}{kA}$ in K/W. Successive layers: resistances in series, $R_{t}=\sum R_{i}$ .
• Convection (a surface bathed by a fluid): $\Phi=hA\,(T_{s}-T_{f})$ . A cooling body: $T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}$ .
• Radiation (even in a vacuum): $P=\varepsilon\sigma AT^{4}$ , net exchange $P=\varepsilon\sigma A\left(T^{4}-T_{e}^{4}\right)$ , $\sigma=5.67\times 10^{-8}$ W/(m²·K⁴).
• Wien's law: $\lambda_{max}T=2.898\times 10^{-3}$ m·K. Insulated cylinder: $R=\dfrac{\ln(r_{2}/r_{1})}{2\pi kL}$ .

In the bar, the energy passes from neighbour to neighbour without anything moving; in the pot, the fluid itself circulates; between the Sun and the Earth, only radiation crosses the vacuum.

The vocabulary of resistances is the same as in electricity: the temperature difference plays the role of the voltage, the flow that of the current.

The spectrum of thermal radiation

• A hot body emits a continuous spectrum whose peak lies at $\lambda_{max}=\dfrac{2.898\times 10^{-3}}{T}$ : the hotter it is, the further this peak slides toward short wavelengths.
• The cooking element at $1073$ K emits mostly around $2.70$ µm, in the infrared: its red glow is only the visible tail of its spectrum.
• The total power, the area under the curve, grows as $T^{4}$ : from $293$ to $1073$ K, it is multiplied by $180$ .

The hotter curve is higher at every wavelength and its peak moves back toward the visible; at

1073

K, the peak stays at

2.70

µm, far into the infrared.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Putting degrees Celsius in a law with T to the fourth power

the whole question, and a result three times too small

What not to write

“Element at $800$ °C: $P=0.90\times\sigma\times 0.018\times 800^{4}=376$ W.”

What to write

“ $T=800+273=1073$ K, so $P=0.90\times\sigma\times 0.018\times 1073^{4}=1.22\times 10^{3}$ W.”

In kelvins, the element radiates

1.22

kW (green); with the number

800

taken as an absolute temperature,

376

W (red): the fourth power multiplies the error by

3.2

Why: The Stefan-Boltzmann law is written for the ABSOLUTE temperature. For a difference $\Delta T$ (Fourier, Newton, $mc\Delta T$ ), Celsius and kelvins give the same number; for $T^{4}$ , never.

2. Calculating the flow in each layer of a wall with the total difference

the whole question, and a flow twenty times too large

What not to write

“Brick: $\dfrac{30}{0.0139}=2160$ W; insulation: $\dfrac{30}{0.25}=120$ W; total flow $2280$ W.”

What to write

“Layers crossed one after the other: $R_{t}=0.0139+0.25=0.264$ K/W, and $\Phi=\dfrac{30}{0.264}=114$ W, the same in each layer.”

In series, a single flow crosses brick then insulation, and it is the

30

K difference that is divided. Flows add only for walls side by side, like a wall and its window.

Why: In series, it is the FLOW that is shared and the temperature difference that is divided among the layers. Adding flows only makes sense for walls side by side, in parallel, like a wall and a window.

3. Forgetting latent heat when ice melts

the whole question, with an absurd final temperature

What not to write

“ $50$ g of ice at $0$ °C in $250$ g of water at $25$ °C: $T=\dfrac{0.25\times 25+0.05\times 0}{0.30}=20.8$ °C.”

What to write

“ $0.25\times 4186\times(25-T)=0.05\times 3.33\times 10^{5}+0.05\times 4186\times T$ , so $T=7.6$ °C.”

Melting uses

16.7

of the

26.2

kJ the water can give up: only

9.5

remain to warm the

300

g, hence

7.6

°C (green) and not

20.8

(red).

Why: Melting the ice costs $16.7$ kJ before it is even warmed, more than half of what the water can give up on its way back to $0$ °C. Always check first whether all the ice melts.

4. Letting the temperature decay instead of the difference

the whole question, and a model that tends toward 0 °C

What not to write

“Coffee at $90$ °C, $k=0.05$ min⁻¹: after $30$ min, $T=90\,e^{-1.5}=20.1$ °C.”

What to write

“It is the difference from the room that decays: $T=20+70\,e^{-1.5}=35.6$ °C.”

The right curve settles onto the 20 °C line; the wrong curve, dashed, crosses the room temperature and keeps going toward 0 °C, which no coffee ever does.

Why: Newton's law acts on $T-T_{a}$ : the body tends toward the ambient temperature, not toward zero. The wrong model predicts a coffee that would end up colder than the room.

5. Leaving the thickness in centimetres in Fourier's law

2 marks, and a result a hundred times too small

What not to write

“Cooler: $\Phi=\dfrac{0.033\times 0.80\times 30}{3}=0.26$ W.”

What to write

“ $e=3.0$ cm $=0.030$ m: $\Phi=\dfrac{0.033\times 0.80\times 30}{0.030}=26.4$ W.”

Why: The conductivity is in W/(m·K): the thickness must be in metres, the area in square metres. A cooler that let in only a quarter of a watt would keep its ice for months.

6. Forgetting what the body receives from its surroundings

3 marks, and a person who would freeze to death within an hour

What not to write

“The human body ( $1.8$ m², $\varepsilon=0.97$ , skin at $306$ K) loses $\varepsilon\sigma AT^{4}=868$ W by radiation.”

What to write

“In a room at $293$ K, it loses $\varepsilon\sigma A\left(306^{4}-293^{4}\right)=138$ W: it receives almost as much as it emits.”

The skin emits

868

W, but the walls send back

730

: only the net exchange,

138

W (green), counts in the balance.

Why: Every body radiates AND absorbs. When the two temperatures are close, the net exchange is a small difference between two large numbers, and it alone counts in a balance.

7. Believing that doubling the insulation on a pipe halves the flow

2 marks on an interpretation question

What not to write

“The insulation goes from $2.5$ to $5.0$ cm thick: the heat flow lost is divided by two.”

What to write

“ $R=\dfrac{\ln(r_{2}/r_{1})}{2\pi kL}$ : the flow is divided by only $\dfrac{\ln(6.1/1.1)}{\ln(3.6/1.1)}=1.44$ .”

Each centimetre added far from the axis offers more surface and resists less: doubling the sleeve multiplies

R

1.44

(green), not by

2

as for a flat wall (red).

Why: The added layer is farther from the axis, its surface is larger: it resists less than the first one. Proportionality to the thickness holds only for a flat wall.

Which method to choose

Which law, from the form of the question

What the question gives to describe the transfer

A single object calls on several branches: conduction through the stopper, radiation between the silvered walls, and the vacuum that removes conduction and convection.

If a thickness, a conductivity k, a solid material → conduction: $\Phi=\dfrac{kA\Delta T}{e}$ , resistances in series for several layers

Example: $\dfrac{0.033\times 0.80\times 30}{0.030}=26.4$ W
If a coefficient h, a fluid in contact with a surface → convection: $\Phi=hA\Delta T$

Example: $25\times 0.054\times 40=54$ W
If an emissivity, a high temperature, a vacuum → radiation: $\varepsilon\sigma A\left(T^{4}-T_{e}^{4}\right)$ , in kelvins

Example: $0.97\sigma\times 1.8\times(306^{4}-293^{4})=138$ W
If a temperature that changes in time toward that of the surroundings → Newton's law: $T=T_{a}+(T_{0}-T_{a})e^{-kt}$

Example: $20+70e^{-1.5}=35.6$ °C
If ice, steam, a plateau → latent heat $Q=mL$ before any $mc\Delta T$

Example: $0.050\times 3.33\times 10^{5}=16.7$ kJ

One problem often combines two branches: the cooler is a conduction that feeds a melting, the vacuum flask a conduction through the stopper plus a radiation between the walls.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

A calorimetry problem with ice, written for the marker

When to use it: The question mixes ice (or steam) with a liquid and asks for a final temperature.

1 Calculate the energy the hot body can give up on its way down to $0$ °C, and the energy needed to bring the ice to $0$ °C and then melt it completely.
2 Compare: if the hot body can supply more than is needed, all the ice melts and the final temperature is above 0 °C; otherwise the final state is a mixture at 0 °C.
3 Write the complete balance, every term with its sign: $m_{w}c_{w}(T_{f}-T_{w})+m_{ice}L_{f}+m_{ice}c_{w}(T_{f}-0)=0$ .
4 Solve, then check that $T_{f}$ lies between $0$ °C and the initial temperature of the hot body.

Concluding sentence

“The water can give up $26.2$ kJ on its way back to $0$ °C, melting requires only $16.7$ : all the ice melts, and the balance gives $T_{f}=7.6$ °C.”

The trap: Writing the balance straight away while assuming all the ice melts, without checking: if the water is not warm enough, you get a negative temperature, a sign that ice remains.

Marking: Typically 1 mark for the comparison of the energies, 2 for the complete balance, 1 for the solution, 1 for the check.

Finding the constant k in Newton's law

When to use it: The question gives two temperature measurements at two instants and the ambient temperature.

1 Write the two DIFFERENCES from the ambient temperature, never the temperatures themselves.
2 Form their ratio: $\dfrac{T_{2}-T_{a}}{T_{1}-T_{a}}=e^{-k(t_{2}-t_{1})}$ .
3 Isolate $k=\dfrac{1}{t_{2}-t_{1}}\ln\dfrac{T_{1}-T_{a}}{T_{2}-T_{a}}$ , with its unit (s⁻¹, min⁻¹ or h⁻¹).
4 Reuse the same ratio to go back or forward in time.

Concluding sentence

“Between the two measurements, the difference goes from $11.0$ to $8.5$ K in $2$ h, so $k=\dfrac{1}{2}\ln\dfrac{11.0}{8.5}=0.129$ h⁻¹.”

The trap: Writing the ratio of the temperatures 29.0 over 26.5 instead of the ratio of the differences: you get a constant almost three times too small.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The final temperature, bracketed

After a calorimetry problem, the equilibrium temperature always lies between the lowest and the highest of the initial temperatures, and closer to that of the body with the greater heat capacity.

Aluminum at $95$ °C in water at $18$ °C: $27$ °C, close to the water, seven times more “capacitive”. A result of $60$ °C is a sign of an error.
Kelvins under the fourth power

Whenever a T appears to the fourth power, reread the numerical value: it must exceed 200.

$800^{4}$ instead of $1073^{4}$ divides the power by $3.2$ .
The flow common to all layers

After a wall calculation, recompute the temperature drop of each layer with the flow you found: their sum must give back the total difference.

$0.66+27.5+1.83=30$ K for the wall of the worked example.
The physical sense of the limit

A temperature that evolves must tend toward that of the surroundings, never overshoot it or tend toward 0 °C.

A coffee “at $15$ °C” in a room at $20$ °C after an hour: wrong model.

The typical problem, taken apart

A three-layer wall: flow and interface temperature

A wall of $10$ m² is made, from inside to outside, of $1.5$ cm of plaster ( $k=0.25$ W/(m·K)), $10$ cm of mineral wool ( $k=0.040$ ) and $2.0$ cm of wood ( $k=0.12$ ). It is $20$ °C inside and $-10$ °C outside.

Calculate the heat flow, then the temperature between the wool and the wood.

Three layers crossed one after the other by the same flow, from

20

°C to

-10

°C: thicknesses to scale and the conductivity of each, nothing calculated.

Step 1

Resistances: $R_{p}=\dfrac{0.015}{0.25\times 10}=0.0060$ ; $R_{m}=\dfrac{0.10}{0.040\times 10}=0.25$ ; $R_{w}=\dfrac{0.020}{0.12\times 10}=0.0167$ K/W.

Why

Convert all thicknesses to metres before dividing: the factor-of-a-hundred error is eliminated on the very first line.

Step 2

In series: $R_{t}=0.0060+0.25+0.0167=0.273$ K/W.

Why

The three layers are crossed one after the other by the same heat: it is the resistances that add up, not the flows.

Step 3

$\Phi=\dfrac{20-(-10)}{0.273}=110$ W.

Why

The difference is 30 K; the double sign is the second trap of the question.

Step 4

Drop in the plaster: $110\times 0.0060=0.66$ K; in the wool: $110\times 0.25=27.5$ K. Wool-wood interface: $20-0.66-27.5=-8.2$ °C.

Why

Each layer uses up a difference $\Phi R$ , as a resistor uses up a voltage: we step down layer by layer from the inside.

Step 5

Check: $-8.2-(-10)=1.8$ K remain for the wood, and $110\times 0.0167=1.8$ K.

Why

The sum of the drops gives back the total difference: the check costs one line and confirms the whole calculation.

The conclusion, written out

“The wall loses $110$ W. The wool carries $92$ % of the resistance: the wool-wood interface is already at $-8.2$ °C, less than two degrees from the outside.”

The classic mistake on this problem: Calculating a flow per layer with the total difference of 30 K and adding them, which gives several kilowatts for an insulated wall.

Learn by heart

• Temperature: a state. Heat: a transfer, from hot to cold.
• $Q=mc\Delta T$ without a change of state; $Q=mL$ during a plateau, at constant temperature.
• Fourier: $\Phi=\dfrac{\Delta T}{R}$ with $R=\dfrac{e}{kA}$ , thickness in metres; in series, $R_{t}=\sum R_{i}$ .
• Newton: it is the difference $T-T_{a}$ that decays as $e^{-kt}$ .
• Stefan-Boltzmann: $P=\varepsilon\sigma AT^{4}$ , $T$ in kelvins; net balance $\varepsilon\sigma A(T^{4}-T_{e}^{4})$ .
• Wien: $\lambda_{max}T=2.898\times 10^{-3}$ m·K.
• Pipe: $R=\dfrac{\ln(r_{2}/r_{1})}{2\pi kL}$ , doubling the insulation does not halve the flow.

Frequently asked questions

What is the difference between heat and temperature?

Temperature describes the state of a body, the average agitation of its particles. Heat is energy that passes from one body to another because their temperatures differ. A body has a thermal energy and a temperature, but never a quantity of heat: heat exists only during the transfer.

Why does the temperature stay constant while ice melts?

Because the energy received goes into breaking the bonds that hold the molecules in the crystal, not into increasing their agitation. As long as some ice remains, the mixture of water and ice stays at zero degrees Celsius. This energy is calculated with the latent heat of fusion, multiplied by the mass that melts.

How do you calculate the heat flow through a wall of several layers?

You calculate the thermal resistance of each layer, the thickness divided by the product of the conductivity and the area, then add these resistances because the heat crosses the layers one after the other. The flow equals the total temperature difference divided by the total resistance, and it is the same in every layer.

Should temperatures be in kelvins in heat transfer calculations?

For a temperature difference, in Fourier's law, Newton's law or the calculation of sensible heat, degrees Celsius and kelvins give the same number. For radiation, where the temperature is raised to the fourth power, the absolute temperature in kelvins is mandatory.

Practise it

Corrected exercises: Heat, temperature and heat transfer, 203-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Special relativity Next sheet Greenhouse effect, albedo and radiative forcing

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys3-heat-transfer. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a 203-SN3-RE tutor in Montreal?

Get in touch for a first session. Heat transfer is worked as a method: identify the mode, write the right law, check the units and the physical sense of the result.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: heat, temperature and heat transfer (203-SN3-RE)

The essentials

Temperature, thermal energy and heat

The three modes and their laws

The spectrum of thermal radiation

The mistakes that cost marks

1. Putting degrees Celsius in a law with T to the fourth power

2. Calculating the flow in each layer of a wall with the total difference

3. Forgetting latent heat when ice melts

4. Letting the temperature decay instead of the difference

5. Leaving the thickness in centimetres in Fourier's law

6. Forgetting what the body receives from its surroundings

7. Believing that doubling the insulation on a pipe halves the flow

Which method to choose

Which law, from the form of the question

How the answer is expected to be written

A calorimetry problem with ice, written for the marker

Finding the constant k in Newton's law

Check before you hand in

The typical problem, taken apart

A three-layer wall: flow and interface temperature

Learn by heart

Frequently asked questions

What is the difference between heat and temperature?

Why does the temperature stay constant while ice melts?

How do you calculate the heat flow through a wall of several layers?

Should temperatures be in kelvins in heat transfer calculations?

Corrected exercises: Heat, temperature and heat transfer, 203-SN3-RE

See also

Looking for a 203-SN3-RE tutor in Montreal?