Heat, Materials and Expansion: Lesson Test

Answer: B

Correct answer: Energy that flows between objects because of a temperature difference. Heat is energy that moves from a hotter object to a cooler one because their temperatures differ. Temperature, by contrast, reflects the average motion of particles and is measured in degrees or kelvins.

• A: That describes what temperature reflects, not heat itself.
• B (correct): This is the definition of heat: energy in transit driven by a temperature difference, measured in joules.
• C: Heat is a form of energy being transferred, not a material substance.
• D: A reading in degrees is a temperature, not an amount of energy transferred.

Answer: D

Correct answer: 313.15 K. Use

K.

• A: 104 is the Fahrenheit value of 40°C, not the Kelvin value.
• B: Kelvin and Celsius degrees are the same size, but the Kelvin scale starts 273.15 degrees lower, so the number changes.
• C: This subtracts 40 from 273.15 instead of adding 273.15 to 40.
• D (correct): Adding 273.15 to the Celsius reading gives 40 + 273.15 = 313.15 K.

Answer: B

Correct answer: Radiation. Space between the Sun and the satellite is nearly a vacuum, so no material can carry heat by contact or by flowing. Only radiation, which needs no medium, can deliver the energy.

• A: Conduction requires direct contact through a material, and there is no material linking the Sun to the satellite.
• B (correct): Radiation travels as electromagnetic waves and is the only one of the three mechanisms that can cross empty space.
• C: Convection needs a moving fluid to carry the energy, and space is essentially a vacuum.
• D: Compression is not a heat transfer mechanism; the three mechanisms are conduction, convection and radiation.

Answer: B

Correct answer: It increases by the same fraction as the plate's other dimensions. Heating a plate uniformly is like enlarging a photograph of it: every length grows by the same fraction. The hole's diameter therefore gets larger, not smaller.

• A: The hole's size is set by the surrounding metal, and that metal's dimensions all grow.
• B (correct): Even heating enlarges every distance in the plate by the same fraction, including the distance across the hole.
• C: This is the common misconception; the metal does not crowd into the hole, because all dimensions scale up together.
• D: The hole grows by the same fraction as other lengths; twice the fractional change applies to area, not to a diameter.

Answer: A

Correct answer: It bends so that the steel side is on the inside of the curve. The high-expansion brass becomes longer than the steel it is bonded to. The longer layer ends up on the outside of the curve, so the strip curls toward the steel side.

• A (correct): The brass side grows longer than the steel side, so the strip curls toward the low-expansion steel.
• B: The two metals grow by different amounts, and because they are bonded the difference forces the strip to bend.
• C: Both metals expand when heated, so neither shrinks, and the unequal growth causes bending.
• D: The longer side must be on the outside of a curve, and the brass becomes the longer side when heated.

Answer: A

Correct answer: To fill tiny air gaps between the surfaces so heat conducts across the joint better. Two metal surfaces that look flat touch only at tiny high points, leaving pockets of air, which conducts heat poorly. Thermal paste fills those pockets with a better conductor, lowering the resistance of the joint.

• A (correct): The surfaces touch only at small high spots, and the paste replaces poorly conducting air in the gaps between them.
• B: Holding the parts together is done by clips or screws; the paste is there to improve heat flow, not as an adhesive.
• C: Insulating the chip would trap heat and cause overheating, the opposite of what the paste is for.
• D: The paste works by conduction across the contact, not by reflecting radiation.

Answer: D

Correct answer: Cast iron. A brittle material breaks with little warning deformation. Cast iron, like glass and ceramics, is brittle, while copper, mild steel and aluminum are ductile.

• A: Mild steel is ductile and usually bends or necks visibly before it fails.
• B: Aluminum is ductile and deforms noticeably before breaking.
• C: Copper is ductile and stretches a great deal before it breaks.
• D (correct): Cast iron is a brittle material that fractures with little or no deformation.

Answer: C

Correct answer: Creep. Creep is gradual, permanent deformation under a steady load that continues over time. It is small at room temperature but significant in hot parts such as turbine blades and boiler tubes.

• A: Fatigue is caused by loads applied and removed many times and leads to cracking, not steady stretching under a constant load.
• B: Elastic deformation disappears when the load is removed, but this stretching is permanent.
• C (correct): Creep is slow, permanent deformation under a sustained load, and it becomes important at high temperature.
• D: Thermal shock is cracking from sudden, uneven heating or cooling, not gradual stretching over time.

Answer: C

Correct answer: It raises the coolant's boiling point so it stays liquid above 100°C. A liquid boils when its vapor pressure equals the pressure above it. Raising the pressure therefore raises the boiling point, so pressurized coolant stays liquid well above 100°C.

• A: Flow is driven by the water pump, and the radiator is still needed to release the heat to the air.
• B: Pressure in this range does not meaningfully change specific heat; its effect here is on the boiling point.
• C (correct): Higher pressure above a liquid raises its boiling point, letting the coolant run hotter without boiling.
• D: Pressure raises, not lowers, the boiling point, and boiling coolant would cool the engine poorly.

Answer: D

Correct answer: Fatigue. Fatigue is failure under many repeated load cycles, each below the stress that would break the part at once. A tiny crack starts at a flaw and grows with every cycle until the remaining metal gives way, which fits a bracket on a vibrating pump.

• A: The loads were all well below the breaking stress, so no single overload caused the failure.
• B: Creep is slow stretching under a steady load, mostly at high temperature, not cracking from repeated loading.
• C: Thermal shock comes from sudden uneven heating or cooling, not from mechanical vibration.
• D (correct): Loads applied and removed many times let a small crack grow cycle by cycle until the part fails, even below its one-pull breaking stress.

Answer: C

Correct answer: 68°F. Use

• A: This adds 32 before multiplying by 9/5, which is the wrong order.
• B: This multiplies by 5/9 instead of 9/5 before adding 32.
• C (correct): Multiplying 20 by 9/5 gives 36, and adding 32 gives 68°F.
• D: This multiplies by 9/5 but forgets to add the 32-degree offset.

Answer: C

Correct answer: 45 Fahrenheit degrees. 100 Celsius degrees span 180 Fahrenheit degrees, so each Celsius degree equals 9/5 Fahrenheit degrees.

For a change, skip the 32:

Fahrenheit degrees.

• A: This adds 32, which is only correct when converting a temperature reading, not a temperature change.
• B: This simply adds 32 to 25, mixing a reading conversion with a change.
• C (correct): A change of 25 Celsius degrees times 9/5 gives 45 Fahrenheit degrees; no offset is added for a change.
• D: This multiplies by 5/9; Fahrenheit degrees are smaller, so the number should get larger.

Answer: A

Correct answer: 15 mm.

m = 15 mm.

• A (correct): $\Delta L=(12\times 10^{-6})(50)(25)=0.015$ m, which is 15 mm.
• B: This multiplies only the coefficient and length and leaves out the 25°C temperature change.
• C: This is a power-of-ten slip when converting 0.015 m to millimeters.
• D: This is a power-of-ten slip in the other direction; 0.015 m is 15 mm, not 150 mm.

Answer: C

Correct answer: It shrinks by 4.6 mm. Cooling is a negative $\Delta T$, so the rail contracts.

a decrease of 4.6 mm.

• A: This is a power-of-ten slip; 0.0046 m is 4.6 mm.
• B: This uses the coefficient for steel instead of the aluminum coefficient given.
• C (correct): $(23\times 10^{-6})(4)(50)=0.0046$ m, and cooling makes the change a decrease.
• D: The size of the change is right, but cooling causes contraction, not expansion.

Answer: A

Correct answer: About 0.19 percent. The fractional length change is

Area grows about twice as fast:

about 0.19 percent.

• A (correct): Area grows by about $2\alpha \Delta T=2(12\times 10^{-6})(80)=0.00192$, or about 0.19 percent.
• B: This is a power-of-ten slip when converting the fraction 0.00192 to a percent.
• C: This is the percent change in length; area grows in two directions, so it is about twice this.
• D: This uses three times the linear coefficient, which applies to volume, not area.

Answer: B

Correct answer: About 69 × 10⁻⁶ per °C. A block grows in length, width and height, so

per °C.

• A: This is the linear coefficient itself; volume grows in three directions, so it changes faster.
• B (correct): A solid expands in three directions, so $\beta \approx 3\alpha =3\times 23\times 10^{-6}$.
• C: Twice the linear coefficient describes area expansion, not volume expansion.
• D: This divides by 3 instead of multiplying; volume expansion is larger than linear expansion.

Answer: D

Correct answer: 72 MPa. For a fully restrained bar,

Pa = 72 MPa.

• A: This drops a power of ten in the product of the modulus and coefficient.
• B: This multiplies only the modulus and coefficient and leaves out the 30°C temperature rise.
• C: This adds an extra power of ten; the product is 72 million pascals.
• D (correct): $\sigma =E\alpha \Delta T=(200\times 10^9)(12\times 10^{-6})(30)=72\times 10^6$ Pa.

Answer: A

Correct answer: 100 W.

• A (correct): $P=kA\Delta T/d=(0.04)(5)(25)/0.05=100$ W.
• B: This multiplies by the thickness instead of dividing by it.
• C: This leaves out the 25 K temperature difference.
• D: This uses 0.005 m for the thickness, a power-of-ten slip.

Answer: C

Correct answer: It is 2/3 of the original. From

heat flow is proportional to $A/d$.

Doubling A and tripling d gives a factor of $2/3$.

• A: This multiplies by both factors, but thicker insulation reduces heat flow instead of increasing it.
• B: This divides by both factors, but a larger area increases heat flow.
• C (correct): Heat flow is proportional to area and inversely proportional to thickness, so it changes by 2/3.
• D: This inverts the ratio, treating thickness as if it increased the heat flow.

Answer: A

Correct answer: It replaces warmed air near the fins with cooler air, keeping the temperature difference large. Still air next to a hot surface warms up and slows further heat transfer. The fan sweeps that air away and brings in cooler air, which is forced convection and greatly increases the rate of heat removal.

• A (correct): This is forced convection: moving air carries heat away and keeps a large temperature difference at the fin surface.
• B: Radiation depends on surface temperature and finish, not on air blowing past the fins.
• C: The transistor's heat output is set by its current, and the fan only helps remove that heat.
• D: Air flow does not change the conductivity of the metal; it changes how fast heat leaves the surface.

Answer: D

Correct answer: About 5.1 times. Radiated power is proportional to $T^4$ with T in kelvins.

so about 5.1 times as much power.

• A: This cubes the temperature ratio instead of raising it to the fourth power.
• B: This squares the temperature ratio instead of raising it to the fourth power.
• C: This assumes radiated power is proportional to temperature instead of its fourth power.
• D (correct): Radiated power scales with $T^4$, and $(450/300{)}^4=1.5^4\approx 5.06$.

Answer: B

Correct answer: 210,000 J. The temperature change is

which equals 25 K.

• A: This uses the final temperature, 45, instead of the temperature change of 25 degrees.
• B (correct): $Q=(2)(4,200)(25)=210,000$ J.
• C: This drops a factor of ten somewhere in the multiplication.
• D: This leaves out the mass of 2 kg.

Answer: B

Correct answer: 1,350 J/K. Heat capacity of an object is

J/K, the energy to raise the whole skillet by one degree.

• A: This is a power-of-ten slip in the product 3 times 450.
• B (correct): Heat capacity is mass times specific heat: (3)(450) = 1,350 J/K.
• C: This is the specific heat per kilogram, not the heat capacity of the whole 3 kg skillet.
• D: This divides the specific heat by the mass instead of multiplying.

Answer: D

Correct answer: 600,000 J. Convert time to seconds: 5 min = 300 s.

• A: This divides power by time instead of multiplying.
• B: This multiplies by 5 without converting minutes to seconds.
• C: This uses 30 s instead of 300 s for five minutes.
• D (correct): Five minutes is 300 s, and (2,000 W)(300 s) = 600,000 J.

Answer: A

Correct answer: 225,600 J. Add the two steps.

Total:

• A (correct): Melting takes 0.6 times 334,000 = 200,400 J, and warming takes 0.6 times 4,200 times 10 = 25,200 J, for 225,600 J in all.
• B: This covers only the melting and leaves out the heat to warm the water by 10 degrees.
• C: This uses the latent heat of vaporization, but melting ice needs the latent heat of fusion.
• D: This covers only the warming and leaves out the large latent heat needed to melt the ice.

Answer: C

Correct answer: 34°C. With the same specific heat, the final temperature is a mass-weighted average.

• A: This weights the masses the wrong way round, giving the hot water the larger weight.
• B: This is the simple average, which applies only when the masses are equal.
• C (correct): The mass-weighted average is (1 × 90 + 4 × 20)/5 = 170/5 = 34°C.
• D: This is just the difference between the two starting temperatures, not a final temperature.

Answer: A

Correct answer: 0.0006. Strain is

with both lengths in the same unit.

with no units.

• A (correct): Converting 1.5 mm to 0.0015 m and dividing by 2.5 m gives 0.0006.
• B: This is a power-of-ten slip in converting millimeters to meters.
• C: This divides millimeters by meters without converting to the same unit.
• D: This divides the original length by the stretch, which is the ratio upside down.

Answer: D

Correct answer: 15 MPa.

Pa = 15 MPa.

• A: This is a power-of-ten slip when dividing by 0.0008.
• B: This multiplies force by area instead of dividing force by area.
• C: This is a power-of-ten slip in the other direction.
• D (correct): $\sigma =F/A=12,000/0.0008=15,000,000$ Pa, or 15 MPa.

Answer: B

Correct answer: 70 GPa.

Pa = 70 GPa, typical of aluminum.

• A: This is a power-of-ten slip in the division.
• B (correct): $E=\sigma /\epsilon =140\times 10^6/0.002=70\times 10^9$ Pa.
• C: This multiplies stress by strain instead of dividing stress by strain.
• D: This is a power-of-ten slip in the other direction; no common metal is that stiff.

Answer: A

Correct answer: To reduce brittleness and restore some toughness. Quenching hardens steel but leaves it brittle. Tempering, a gentle reheat afterward, reduces that brittleness and restores some toughness so the tool resists chipping.

• A (correct): This gentle reheat is tempering, which relieves the brittleness left by quenching.
• B: Hardening was done by the quench; tempering trades a little hardness for less brittleness.
• C: Making the steel soft for shaping is annealing, which uses slow cooling from high temperature.
• D: Tempering is used to adjust mechanical properties, not to improve heat conduction.

Answer: C

Correct answer: 400,000 J.

At 30 m/s:

At 10 m/s:

The brakes absorb

• A: This uses the speed change of 20 m/s in the energy formula; energy depends on each speed squared, not on the difference.
• B: This is the full starting kinetic energy, as if the car stopped completely.
• C (correct): The kinetic energy drops from 450,000 J to 50,000 J, a loss of 400,000 J that becomes heat.
• D: This leaves out the one-half in the kinetic energy formula.

Answer: A

Correct answer: About 16 times. Convert to kelvins: 27°C ≈ 300 K and 327°C ≈ 600 K, so the absolute temperature doubles.

Radiated power scales with $T^4$, so it grows by

• A (correct): In kelvins the temperature goes from 300 K to 600 K, a doubling, and $2^4=16$.
• B: This raises the Celsius ratio 327/27 to the fourth power; the formula requires kelvins.
• C: This uses the kelvin ratio correctly but forgets that power depends on the fourth power of temperature.
• D: This uses the Celsius ratio and ignores the fourth-power dependence.

Answer: C

Correct answer: The two stresses are equal. For a fully restrained bar,

which does not include length.

Bar 1:

Bar 2:

With the same $\Delta T$, the stresses are equal.

• A: This looks only at the stiffness and ignores bar 2's smaller expansion coefficient.
• B: This looks only at the expansion coefficients and ignores the difference in stiffness.
• C (correct): Restrained thermal stress depends on $E\alpha \Delta T$; both bars have $E\alpha =2\times 10^6$ Pa per °C, and length does not matter.
• D: This assumes stress grows with length, but restrained thermal stress does not depend on length.

Answer: C

Correct answer: 2.3 mm. First find the temperature rise:

Then

m = 2.3 mm.

• A: This is a power-of-ten slip when converting 0.0023 m to millimeters.
• B: This is a power-of-ten slip in the other direction.
• C (correct): The heat raises the rod by 54,000/(1.2 × 900) = 50°C, and $(23\times 10^{-6})(2)(50)=0.0023$ m.
• D: This finds the temperature rise as 54,000/900 = 60°C, leaving out the rod's mass.

Answer: D

Correct answer: About 25°C. Heat lost by iron = heat gained by water:

The water's much larger heat capacity keeps the result close to 20°C.

• A: This gives the iron's heat capacity to the water and the water's to the iron, reversing the weighting.
• B: This averages by mass only and ignores the large difference in specific heats.
• C: This is the simple average of the two temperatures, ignoring both mass and specific heat.
• D (correct): Setting heat lost by the iron equal to heat gained by the water gives $T_f=213,000/8,625\approx 24.7$°C.

Answer: D

Correct answer: 1,256,000 J. Warming:

Boiling:

Total:

• A: This uses 100 degrees as the temperature change instead of 100 − 40 = 60 degrees.
• B: This includes only the warming step and leaves out the large latent heat of vaporization.
• C: This includes only the boiling step and leaves out warming the water from 40°C to 100°C.
• D (correct): Warming takes 126,000 J and boiling takes 1,130,000 J, for a total of 1,256,000 J.

Answer: B

Correct answer: About 315 s. Heat for the water:

Energy drawn at 80 percent efficiency:

• A: This divides the 504,000 J the water needs by 2,000 W and ignores the 20 percent of energy lost.
• B (correct): The water needs 504,000 J; at 80 percent efficiency the kettle must draw 630,000 J, and 630,000 divided by 2,000 W is 315 s.
• C: This applies the 80 percent efficiency twice, overstating the energy the kettle must draw.
• D: This multiplies by 0.80 instead of dividing, which wrongly makes the losses shorten the heating time.

Answer: D

Correct answer: 100°C. The bore diameter behaves like any other length, so use

• A: This is a power-of-ten slip in the division.
• B: This uses twice the linear coefficient, as for area; a diameter is a length, so the linear coefficient applies.
• C: This is a power-of-ten slip in the other direction, far hotter than a shrink fit needs.
• D (correct): $\Delta T=\Delta L/(\alpha L_0)=0.06/(12\times 10^{-6}\times 50)=100$°C.

Answer: B

Correct answer: Bar X stretches less, because it has the higher Young modulus. Same load and same area give the same stress in both bars. Strain is $\sigma /E$, so the bar with the larger modulus stretches less.

Steel's 200 GPa beats aluminum's 70 GPa, so the steel bar stretches about a third as much, regardless of which bar is stronger.

• A: Thermal expansion has nothing to do with stretch under a mechanical load at constant temperature.
• B (correct): How much a bar stretches under a given stress is set by stiffness, and steel's modulus is about three times aluminum's.
• C: This confuses strength with stiffness; yield strength tells when permanent deformation starts, not how much the bar stretches elastically.
• D: Equal stress does not mean equal strain; strain equals stress divided by each material's modulus.

Answer: B

Correct answer: The water, about 31 times as much. Bolt:

Water:

840,000/27,000 ≈ 31, so the cooler but much larger, high-specific-heat water holds far more energy.

• A: This compares only the temperatures, 300/40, ignoring mass and specific heat.
• B (correct): The water releases 840,000 J and the bolt releases 27,000 J; 840,000/27,000 is about 31.
• C: Temperature alone does not set stored thermal energy; mass and specific heat matter too.
• D: This compares only the specific heats, 4,200/450, and leaves out the masses and temperature drops.

Answer: A

Correct answer: About 2.1 mm. Use the temperature change, not a single reading: ΔT =

That is about 2.1 mm.

• A (correct): The temperature change is 50°C, and 17 × 10⁻⁶ × 2.5 × 50 = 0.002125 m, about 2.1 mm.
• B: This uses the final temperature, 70°C, as the change; only the 50 degree rise counts.
• C: This uses the starting temperature, 20°C, as the change instead of the 50 degree rise.
• D: This slips a power of ten when converting 0.002125 m to millimeters.

Answer: A

Correct answer: 1.0 mm. Find strain from Young's modulus, then multiply by the length.

Stretch:

m = 1.0 mm.

• A (correct): Strain is 100 MPa ÷ 200 GPa = 0.0005, and 0.0005 × 2 m = 0.001 m, or 1.0 mm.
• B: This is the strain read as millimeters per meter but not multiplied by the 2 m length.
• C: This divides the 0.0005 strain by the 2 m length; stretch is strain multiplied by length.
• D: This slips a power of ten when combining megapascals and gigapascals.

Answer: D

Correct answer: 60°C. Both bars expand into the gap, so add their growth per degree:

m per °C.

Starting from 20°C, the ends touch at about 60°C.

• A: This is the temperature rise needed; it must be added to the starting 20°C to give the closing temperature.
• B: This counts only the aluminum bar's growth; the steel bar also expands into the gap.
• C: This counts only the steel bar's growth; the aluminum bar also expands into the gap.
• D (correct): Together the bars grow 70 × 10⁻⁶ m per degree, so closing 0.0028 m takes a 40 degree rise, from 20°C to 60°C.

Answer: C

Correct answer: About 0.94 kg. Heat lost by the lemonade equals heat gained by the melting ice.

So

• A: This divides by the latent heat of vaporization; melting ice uses the latent heat of fusion.
• B: This uses 298 K as the temperature change; a drop from 25°C to 0°C is a change of only 25 K.
• C (correct): Cooling the lemonade releases 3 × 4,200 × 25 = 315,000 J, which melts 315,000 ÷ 334,000 ≈ 0.94 kg of ice.
• D: This divides the latent heat by the heat released, flipping the ratio.

Answer: C

Correct answer: A low coefficient of thermal expansion. Thermal shock cracks a brittle material when one region heats or cools faster than another and the two try to change size by different amounts. A material with low expansion develops much less of this internal stress, so it resists cracking.

• A: Specific heat sets how much energy a temperature change takes; it does not limit the stress from uneven expansion.
• B: Thermal shock cracking happens far below melting; it comes from uneven expansion, not from softening.
• C (correct): With little expansion, hot and cool regions try to change size by only slightly different amounts, so internal stress stays small.
• D: Hardness resists scratches and dents; it does not reduce stress from uneven heating.

Answer: B

Correct answer: It reflects radiated body heat back and emits little heat itself. Dark, dull surfaces absorb and emit radiation well, while shiny surfaces reflect it and emit poorly. The metal coating reflects infrared radiation from the body back toward it and radiates little energy outward.

• A: Fast conduction would carry heat away from the body, the opposite of what the blanket is for.
• B (correct): Shiny, light surfaces reflect most radiation and are poor emitters, so body heat is held in.
• C: Shiny surfaces reflect radiation rather than absorb it, and no phase change takes place.
• D: Speeding up convection would remove heat faster; the coating is meant to reduce heat loss.

Answer: B

Correct answer: 30 percent. A watt is one joule per second, so each second the generator delivers 2,400 J of useful energy from 8,000 J of fuel energy.

• A: This is the share of the input lost as waste heat, not the useful share.
• B (correct): Efficiency is useful output over input: 2,400 J ÷ 8,000 J = 0.30, or 30 percent.
• C: This divides input by output; an efficiency above 100 percent signals the ratio is upside down.
• D: This misplaces the decimal when changing 0.30 into a percent.

Answer: D

Correct answer: Annealing. Annealing means heating a metal and cooling it slowly. It softens the metal and makes it easier to shape, the opposite of quenching, which hardens it.

• A: Quenching is rapid cooling in water or oil, which makes steel harder and more brittle, not softer.
• B: Tempering is a gentle reheat after quenching to reduce brittleness; it does not involve slow cooling from high heat to soften fully.
• C: Creep is slow permanent deformation under a steady load at high temperature, not a heat treatment.
• D (correct): Slow cooling from a high temperature softens metal and makes it easier to form.

Answer: D

Correct answer: 22,500 J. Use the temperature change, which is the same in Celsius degrees and kelvins:

K.

• A: This uses the final temperature, 120°C, as the change instead of the 100 degree rise.
• B: This converts 120°C to 393 K and uses it as the change; a temperature change in kelvins equals the change in Celsius degrees.
• C: This uses the starting temperature, 20°C, as the change.
• D (correct): The temperature change is 100 K, and 0.5 × 450 × 100 = 22,500 J.

Answer: B

Correct answer: Natural convection, as warmed air rises and cooler air sinks. In natural convection, a fluid warmed at a heat source becomes less dense and rises, while cooler fluid sinks to take its place. That circulating current carries heat from the floor-level heater throughout the room.

• A: Air is a poor conductor; heat spreading through a whole room depends on the air moving.
• B (correct): Air heated at the floor becomes less dense and rises while cooler air sinks to replace it, setting up a circulating current.
• C: Forced convection needs a fan or pump; this heater has none, so the flow is driven by density differences.
• D: Radiation from a low heater warms nearby surfaces, but it is the moving air that carries heat up to the ceiling.