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

$Q=mc\Delta T=1.50\times 4186\times 80.0=5.02\times 10^{5}$ J, or $502$ kJ. The mass comes from the volume: $1.50$ L of water has a mass of $1.50$ kg.

$\Delta T=80.0$ K. A difference of one degree Celsius is a difference of one kelvin: the two scales differ only in their zero. Only temperatures themselves are converted to kelvins (by adding $273.15$), never differences, and it is an absolute temperature that the laws of radiation will require.

$t=\dfrac{Q}{P}=\dfrac{5.02\times 10^{5}}{2000}=251$ s, or $4$ min $11$ s.

The electrical energy consumed is $2000\times 270=5.40\times 10^{5}$ J. Efficiency: $\eta=\dfrac{5.02\times 10^{5}}{5.40\times 10^{5}}=0.930$, or $93.0$ %. The remaining $7$ % heats the vessel, the element and the air of the kitchen.

The statement mixes up two quantities. TEMPERATURE is a state quantity, tied to the average agitation of the particles. THERMAL ENERGY (internal energy) is what the water possesses, and it is greater at $100$ °C. HEAT, on the other hand, is never contained: it is energy IN TRANSIT, exchanged because of a temperature difference. The correct statement is therefore: water at $100$ °C has a greater thermal energy, and it received $502$ kJ of heat to get there.

In an insulated enclosure, the sum of the heats exchanged is zero: $m_{Al}c_{Al}(T_{eq}-95.0)+m_{w}c_{w}(T_{eq}-18.0)=0$. Hence $T_{eq}=\dfrac{m_{Al}c_{Al}\cdot 95.0+m_{w}c_{w}\cdot 18.0}{m_{Al}c_{Al}+m_{w}c_{w}}=\dfrac{225\times 95.0+1674.4\times 18.0}{225+1674.4}=27.1$ °C. Writing every term in the form $mc(T_{f}-T_{i})$, with its sign, prevents mistakes about who gives and who receives.

$Q_{Al}=225\times(27.1-95.0)=-1.53\times 10^{4}$ J: the aluminum gives up $15.3$ kJ, which the water receives.

The actual temperature ($26.9$ °C) is lower than the prediction ($27.1$ °C): part of the energy given up by the aluminum did not go into heating the water. The calorimeter, which started at $18.0$ °C, ABSORBED it by warming up along with the water.

Balance including the calorimeter: $m_{Al}c_{Al}(95.0-26.9)=(m_{w}c_{w}+K)(26.9-18.0)$. So $K=\dfrac{225\times 68.1}{8.9}-1674.4=1721.6-1674.4=47$ J/K. The calorimeter behaves like $11$ g of extra water: this is its “water equivalent”. Note the sensitivity: an error of $0.2$ °C in the reading is enough to make it appear.

Because the heat capacities are not comparable: the water has $m_{w}c_{w}=1674$ J/K, the aluminum only $225$ J/K, more than seven times less. The same energy exchanged therefore produces a temperature change seven times smaller in the water. It is the high specific heat capacity of water that makes it such a good thermal regulator.

In one hour, $Q=700\times 3600=2.52\times 10^{6}$ J. Mass evaporated: $m=\dfrac{Q}{L_{v}}=\dfrac{2.52\times 10^{6}}{2.42\times 10^{6}}=1.04$ kg. A litre of sweat per hour is a realistic order of magnitude when running in hot weather, which is why drinking matters.

It is the CHANGE OF STATE that takes up the energy, not the presence of water on the skin. A drop that falls carries the water away without having vaporized it, and so without having taken the $2.42$ MJ/kg from the skin. In very humid weather, saturated air blocks evaporation and the same phenomenon shuts off the cooling.

Vaporizing: $Q_{v}=1.00\times 2.26\times 10^{6}=2.26\times 10^{6}$ J. Heating: $Q_{c}=1.00\times 4186\times 80=3.35\times 10^{5}$ J. Ratio $\dfrac{2.26\times 10^{6}}{3.35\times 10^{5}}=6.75$. Boiling a pot dry takes almost seven times more energy than bringing it to a boil.

Two steps, which are never merged. Warming the ice from $-10.0$ to $0$ °C: $1.00\times 2090\times 10.0=2.09\times 10^{4}$ J, using the heat capacity of ICE. Melting: $1.00\times 3.33\times 10^{5}$ J. Total: $3.54\times 10^{5}$ J.

The energy received during melting breaks the bonds that hold the molecules in the crystal lattice: it increases the interaction potential energy, not the average agitation. The temperature, which measures this agitation, therefore stays constant over the whole plateau, as long as some ice remains.

$R=\dfrac{e}{kA}=\dfrac{4.0\times 10^{-3}}{0.80\times 1.20}=4.17\times 10^{-3}$ K/W. The thickness must be in metres.

$\Phi=\dfrac{\Delta T}{R}=\dfrac{30}{4.17\times 10^{-3}}=7.2\times 10^{3}$ W. More than seven kilowatts for a single window: the result is absurd, and that is a warning sign. The glass, thin and fairly conductive, is not what limits the flow; part e) will say what does.

The layers are crossed by the SAME flow, one after the other: their resistances are in SERIES and add up. $R_{air}=\dfrac{0.012}{0.026\times 1.20}=0.385$ K/W, so $R_{t}=2\times 4.17\times 10^{-3}+0.385=0.393$ K/W. The air gap provides $98$ % of the resistance.

$\Phi=\dfrac{30}{0.393}=76$ W, about $94$ times less than the calculation in b).

$R=4.17\times 10^{-3}+0.138=0.142$ K/W, and $\Phi=\dfrac{30}{0.142}=2.1\times 10^{2}$ W. It is the still air films on each face, where heat passes only by conduction in a gas, that provide most of the insulation of a single glazing. The glass itself contributes almost nothing.

$T=800+273=1073$ K. In every law of radiation, the temperature is ABSOLUTE: $T^{4}$ in degrees Celsius has no physical meaning.

$P=\varepsilon\sigma AT^{4}=0.90\times 5.67\times 10^{-8}\times 0.018\times 1073^{4}=1.22\times 10^{3}$ W.

The room also radiates toward the element: $P_{net}=\varepsilon\sigma A\left(T^{4}-T_{r}^{4}\right)=1.21\times 10^{3}$ W. The correction is tiny here ($7$ W), because $293^{4}$ is $180$ times smaller than $1073^{4}$; it becomes decisive between two bodies at nearby temperatures.

$\lambda_{max}=\dfrac{2.898\times 10^{-3}}{1073}=2.70\times 10^{-6}$ m $=2.70$ µm: in the INFRARED. The element looks dull red because only the tail of its spectrum, on the short-wavelength side, spills over into the visible.

$T^{4}=\dfrac{1000}{\varepsilon\sigma A}+T_{r}^{4}$, hence $T=1023$ K, or $750$ °C. Losing just $50$ °C cuts the power by $17$ %: that is the $T^{4}$ law; near $1000$ K, $\dfrac{\Delta P}{P}\approx 4\dfrac{\Delta T}{T}$.

$R_{ins}=\dfrac{e}{kA}=\dfrac{0.140}{0.040\times 12.0}=0.292$ K/W.

$R_{g}=\dfrac{0.013}{0.17\times 12.0}=6.37\times 10^{-3}$ K/W and $R_{b}=\dfrac{0.090}{0.72\times 12.0}=1.04\times 10^{-2}$ K/W. Layers crossed one after the other by the same flow are in series: $R_{t}=0.00637+0.292+0.0104=0.308$ K/W.

$\Phi=\dfrac{\Delta T}{R_{t}}=\dfrac{39}{0.308}=126$ W. The difference is $21-(-18)=39$ K: watch the double sign.

This is the analogue of Ohm's law: each layer “uses up” a temperature drop $\Phi R$. Across the gypsum: $126\times 0.00637=0.81$ K, giving $20.2$ °C. Across the insulation: $126\times 0.292=36.9$ K, giving $20.2-36.9=-16.7$ °C at the interface. We check that $1.3$ K remains for the brick, which does bring us back to $-18$ °C.

$\dfrac{0.292}{0.308}=0.946$: $94.6$ % of the resistance, and so $94.6$ % of the temperature drop, takes place in the insulation. Practical consequence: almost everything outside the insulation is at the outdoor temperature, and the vapour barrier must go on the warm side of the insulation, otherwise the water vapour from the house condenses in the cold insulation.

The equation is separable: $\dfrac{dT}{T-T_{a}}=-k\,dt$, so $\ln|T-T_{a}|=-kt+C$ and $T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}$. It is the DIFFERENCE from the ambient air that decays exponentially, not the temperature itself: the body never tends toward $0$ °C but toward $18$ °C.

Between the two measurements, $2$ h go by and the difference goes from $11.0$ to $8.5$ K: $e^{-2k}=\dfrac{8.5}{11.0}$, so $k=\dfrac{1}{2}\ln\dfrac{11.0}{8.5}=0.129$ h⁻¹.

At death the difference was $37.0-18.0=19.0$ K, at 11:00 p.m. it is $11.0$ K: $e^{-kt}=\dfrac{11.0}{19.0}$, so $t=\dfrac{1}{k}\ln\dfrac{19.0}{11.0}=4.24$ h, or $4$ h $14$ min. Death goes back to about 6:46 p.m.

Two hours after 1:00 a.m.: $T=18.0+8.5\,e^{-2k}=18.0+8.5\times\dfrac{8.5}{11.0}=24.6$ °C. The ratio $\dfrac{8.5}{11.0}$ repeats every two hours: that is the signature of the exponential.

Because the assumptions are wrong at the start: the body barely cools during the first hour (post-mortem plateau), clothing and position change $k$, and $k$ itself varies because radiation and evaporation are not linear in $T-T_{a}$. The estimate remains useful as an order of magnitude, with a range of several hours.

The sweater produces no energy: it traps still air between its fibres and increases the thermal RESISTANCE between the skin and the outdoors. It is the body that heats; the sweater reduces the heat flow lost. Placed on a block of ice, the same sweater would slow its melting.

At equilibrium, both are at the temperature of the room. The handle FEELS colder because metal, a very good conductor with a high effusivity, pulls heat out of the hand much faster than wood does: the hand measures a flow, not a temperature.

It is HOT AIR that rises, by convection, because it is less dense. Heat itself spreads in every direction by conduction and by radiation: a radiator radiates toward the floor as much as toward the ceiling.

It is the opposite. A black body ABSORBS all the radiation it receives ($\varepsilon=1$), and that is precisely why it is the ideal emitter: at a given temperature, no body emits more than it does. It looks black at room temperature only because its emission is then in the infrared.

The Stefan-Boltzmann law involves the ABSOLUTE temperature: $\dfrac{P_{2}}{P_{1}}=\left(\dfrac{313.15}{293.15}\right)^{4}=1.30$. The power increases by $30$ %, not by $100$ %. To double it, we would need $T=293.15\times 2^{1/4}=348.6$ K, or $75$ °C.

Conduction and convection need matter: the vacuum eliminates them between the two walls. Radiation, on the other hand, crosses the vacuum (this is how the Sun's energy reaches us): another defence is needed, the silvering.

$\varepsilon_{eff}=\dfrac{1}{2/0.030-1}=0.0152$. $P=5.67\times 10^{-8}\times 0.060\times 0.0152\times\left(363^{4}-293^{4}\right)=0.52$ W.

$\varepsilon_{eff}=\dfrac{1}{2/0.90-1}=0.818$ and $P=27.8$ W, fifty-four times more. The silvering divides the radiation in the ratio of the effective emissivities.

$\Phi=\dfrac{kA\,\Delta T}{e}=\dfrac{0.040\times 1.2\times 10^{-3}\times 70}{0.025}=0.134$ W. The stopper accounts for a fifth of the losses: once the vacuum and the silver are in place, the neck becomes the weak link.

Energy to lose: $0.50\times 4186\times 30=6.28\times 10^{4}$ J. Losses: $0.52+0.134=0.65$ W. Time: $\dfrac{6.28\times 10^{4}}{0.65}=9.6\times 10^{4}$ s, about $27$ h. This is an optimistic bound within its model (the glass neck conducts too, and real flasks keep coffee hot for about twelve hours) but pessimistic on one point: the losses fall as the coffee cools.

$P_{c}=(UA)_{tot}\,\Delta T=(180+45)\times 45=1.01\times 10^{4}$ W. The difference is $20-(-25)=45$ K.

Volume renewed per hour: $0.5\times 350=175$ m³, or $175\times 1.20=210$ kg. Flow rate: $\dot m=\dfrac{210}{3600}=0.0583$ kg/s.

Ventilation is FORCED convection: air at $-25$ °C replaces air at $20$ °C. $P_{v}=\dot m\,c\,\Delta T=0.0583\times 1005\times 45=2.64\times 10^{3}$ W.

$P=10\,125+2638=1.28\times 10^{4}$ W. Over $24$ h: $E=1.276\times 10^{4}\times 86\,400=1.10\times 10^{9}$ J, or $\dfrac{1.10\times 10^{9}}{3.6\times 10^{6}}=306$ kWh.

Only $25$ % of the ventilation losses remain: $P=10\,125+0.25\times 2638=1.08\times 10^{4}$ W. The gain ($2.0$ kW) is real, but the envelope remains the dominant item: it is the insulation, not the ventilation, that decides the bill.

$e_{skin}=\sqrt{0.37\times 1000\times 3500}=1.14\times 10^{3}$ and $e_{cer}=\sqrt{1.3\times 2300\times 840}=1.58\times 10^{3}$ J/(m²·K·s$^{1/2}$). Effusivity combines conduction and heat capacity: it measures how easily a material SWALLOWS a supply of heat at its surface.

$T_{c}=\dfrac{1138\times 33+1585\times 20}{1138+1585}=25.4$ °C. The contact temperature leans toward the body of greater effusivity: the tile almost “imposes” its temperature on the skin.

$e_{wood}=\sqrt{0.12\times 700\times 1700}=378$, so $T_{c}=\dfrac{1138\times 33+378\times 20}{1516}=29.8$ °C.

$e_{rug}=\sqrt{0.060\times 160\times 1300}=112$, so $T_{c}=\dfrac{1138\times 33+112\times 20}{1250}=31.8$ °C, barely more than a degree below the skin.

The receptors in the skin measure their OWN temperature, and so the flow that cools them, never the temperature of the floor. On contact with the tile, the surface of the skin drops from $33$ to $25$ °C: a sensation of cold. On the rug, it loses only $1.2$ °C. A thermometer placed on the three floors would nevertheless read $20$ °C on all of them.

The same energy per second crosses every cylinder, but the surface $2\pi rL$ grows with $r$: the heat spreads over a larger area. The gradient $\left|\dfrac{dT}{dr}\right|=\dfrac{\Phi}{2\pi kLr}$ is therefore steep near the pipe and shallow toward the outside; the temperature is not an affine function of $r$ but a logarithmic one.

$\dfrac{dT}{dr}=-\dfrac{\Phi}{2\pi kL}\cdot\dfrac{1}{r}$, which we integrate: $T_{1}-T_{2}=\dfrac{\Phi}{2\pi kL}\ln\dfrac{r_{2}}{r_{1}}$. So $R=\dfrac{\ln(3.6/1.1)}{2\pi\times 0.035\times 8.0}=\dfrac{1.186}{1.759}=0.674$ K/W. The ratio of the radii is written without conversion, since the units cancel.

$\Phi=\dfrac{70-15}{0.674}=81.6$ W.

$R'=\dfrac{\ln(6.1/1.1)}{1.759}=0.974$ K/W and $\Phi'=\dfrac{55}{0.974}=56.5$ W.

$\dfrac{81.6}{56.5}=1.44$. Doubling the thickness does not divide the flow by two: the resistance grows like the LOGARITHM of the radius, and the added layer, farther from the axis, has a larger surface. On a flat wall, the same operation would have cut the flow in half.

$A=\dfrac{\Phi}{h\,\Delta T}=\dfrac{65}{30\times 40}=5.42\times 10^{-2}$ m², or $542$ cm².

We are short by $542-25=517$ cm², at $2\times 5.0\times 3.0=30$ cm² per fin: $\dfrac{517}{30}=17.2$. We round UP: $18$ fins. Rounding to $17$ would exceed $75$ °C.

The resistances are in series along the path of the heat: the paste adds $\Delta T=\Phi R=65\times 0.20=13$ K, so $T_{chip}=75+13=88$ °C.