Thermo-physics solutions

Convert mass and energy: m = 5000 g; Q = 1.00 × 10⁷ J.

Why: The specific heat is per gram, so convert the block’s mass and supplied energy to compatible units.

First find the energy to reach the first listed threshold: Q_to_threshold = (5000)(0.71)(3550 − 500) = 10,827,500 J.

Why: Before using any phase-change term, calculate the sensible heat needed to reach the first listed threshold.

The available 10,000,000 J is smaller, so the model remains on the solid-heating segment. No latent heat is used.

Why: The supplied energy runs out before that threshold, so all of it raises the solid’s temperature.

ΔT = Q/(mc) = 10,000,000/(5000 × 0.71) = 2816.9014 K. A Celsius temperature difference has the same numerical value as a kelvin difference, so T_f = 500 + 2816.9014 = 3316.9014°C.

Why: Divide the heat by the solid’s heat capacity and add the rise to 500°C; temperature differences have equal Celsius and kelvin sizes.

The water mass is 200 g, and the metal mass is 100,000 g; retain the stated 100 kg rather than silently changing it to 100 g.

Why: Density turns water volume into mass; keep the unusually large printed metal mass so the calculation tests the stated data.

The water gains Q_w = m_wc_w(65 − 15) = (200)(4.18)(50) = 41,800 J.

Why: The water warms by 50 K, so its heat gain follows from its mass, specific heat and that rise.

For an isolated system the metal loses this same energy: m_m c_m(80 − 65) = 41,800 J.

Why: With no heat escaping, the metal’s 15 K cooling must release the same energy the water gains.

Thus c_m = 41,800/(100,000 × 15) = 0.0278667 J/(g·K) = 27.8667 J/(kg·K).

Why: Divide the measured heat transfer by the metal mass and temperature drop to infer its specific heat.

This is atypically low for ordinary metals near these temperatures. The calculation is consistent with the printed numbers, but those numbers should be checked against the actual apparatus; no material identity can be inferred reliably.

Why: The tiny inferred specific heat is a reason to question the source data, rather than quietly replacing its mass.

The heat capacities are C_w = 500(4.18) = 2090 J/K, C_Fe = 5(0.45) = 2.25 J/K and C_Al = 20(0.91) = 18.20 J/K.

Why: Multiply each mass by its own specific heat to see how much energy each object needs per degree.

Write the energy balance with signs: 2090(T_f − 20) + 2.25(T_f − 80) + 18.20(T_f − 80) = 0.

Why: Use signed temperature changes: water gains energy and both warmer metals lose energy in the insulated system.

Collect terms: T_f = [2090(20) + (2.25 + 18.20)(80)]/(2090 + 2.25 + 18.20) = 20.58139°C.

Why: Collect the common final-temperature terms to form a heat-capacity-weighted mean.

The result lies between 20°C and 80°C and stays near the water’s initial temperature because its heat capacity greatly exceeds those of the metals.

Why: Water dominates the total heat capacity, so a final temperature close to 20°C is the expected check.

Use m = 2000 g and Q = 175,000 J. The temperature rise is ΔT = Q/(mc) = 175,000/(2000 × 0.39) = 224.35897 K.

Why: Use the copper’s heat capacity to turn absorbed energy into the temperature rise that drives expansion.

For the supplied constant-coefficient small-expansion model, ΔL = αL_0ΔT = (17 × 10⁻⁶)(0.50)(224.35897) = 0.00190705 m.

Why: Free expansion changes length by α times its original length and temperature rise in the supplied small-strain model.

Add the initial length: L_f = 0.50 + 0.00190705 = 0.50190705 m. The fractional expansion is about 0.00381, within the small-strain approximation.

Why: Add the extension to the starting length and check that the fractional change is small enough for that model.

Cooling liquid water to 0°C requires removal of Q_cool = (200)(4.18)(80) = 66,880 J.

Why: The water must cool to its freezing point before energy removal can freeze it.

Freezing it at 0°C requires an additional Q_freeze = mL_f = (200)(334) = 66,800 J.

Why: At 0°C more energy must leave to change phase, even though the temperature no longer falls.

Total energy removed is 133,680 J. At constant removal power P = 10 J/s, t = Q/P = 13,368 s.

Why: Add cooling and freezing energy, then divide by the device’s constant energy-removal rate.

This is 222.8 min, or 3.7133 h. No further ice cooling is required because the target is just-completed freezing.

Why: Convert seconds to useful time units; stopping at ice at 0°C means no extra ice-cooling stage belongs in the total.

The source picture places the hot and cold faces at opposite ends of the 45 m dimension. Their area is A = 50 × 20 = 1000 m², and L = 45 m.

Why: Conduction uses the area perpendicular to heat flow and the distance between the maintained hot and cold faces.

The maintained temperature difference is ΔT = 300 − 25 = 275 K.

Why: Subtract the two face temperatures; their Celsius difference is the same 275 K gradient-driving difference.

For steady conduction, P = kAΔT/L = (428)(1000)(275)/45 = 2,615,555.56 W.

Why: The steady Fourier-law rate increases with face area and temperature difference and decreases with flow length.

In 10 s the transmitted energy is Q = Pt = 26,155,555.6 J = 26.1556 MJ. This is energy crossing the block in the specified time, not the energy needed to create its initial temperature profile.

Why: Multiply the steady power by 10 s to count transmitted energy, rather than energy stored in the silver.

The density at the specified initial state gives fixed mass m_s = ρ_i V_i = 7.85 g. Initial edge length is L_i = (1.00 cm³)^(1/3) = 1.00 cm.

Why: Use density at the stated initial state to fix the steel’s mass, and take the cube root of volume for its edge.

Heat capacities are C_s = 7.85(0.49) = 3.8465 J/K and C_w = 500(4.18) = 2090 J/K.

Why: Each mc product measures the object’s energy change per degree, with water overwhelmingly larger.

Energy conservation gives T_f = [3.8465(150) + 2090(65)]/(2093.8465) = 65.15615°C.

Why: An insulated energy balance makes the common temperature a heat-capacity-weighted mean of the two starting temperatures.

The steel’s temperature change is ΔT = 65.15615 − 150 = −84.84385 K. Its edge contracts to L_f = L_i[1 + αΔT] = 0.99906672 cm.

Why: The steel cools relative to 150°C, so its negative temperature change contracts each edge.

Using the cubed linear-expansion result gives V_f = L_f³ = 0.99720277 cm³. The first-order volume approximation V_i[1 + 3αΔT] gives 0.99720015 cm³, agreeing to first order. The cube’s mass stays fixed while its density increases.

Why: Cube the final edge to get volume; the fixed mass means this smaller volume has a larger density.

Both start as liquids above their melting points. Each passes through liquid heating, vaporization and gas heating; fusion is not part of either path.

Why: The starting states are already liquid, so include only warming the liquid, boiling it and warming the resulting gas.

For water: Q_w = 6000[4.18(100 − 90) + 2230 + 1.86(400 − 100)] J = 16,978,800 J.

Why: For water, add its three per-gram energy stages and multiply by the 6000 g mass.

For acetone: Q_a = 9000[2.18(56 − 20) + 549 + 1.28(400 − 56)] J = 9,610,200 J.

Why: For acetone, use its own boiling point and three material properties before multiplying by 9000 g.

Subtract: Q_w − Q_a = 7,368,600 J. Acetone requires less energy despite its larger mass, chiefly because the model gives it a much smaller vaporization energy.

Why: Compare the total energies; acetone’s much smaller latent heat outweighs its larger sample mass in this model.

The heat capacities are C_w = 500(4.18) = 2090 J/K and C_Fe = 5(0.45) = 2.25 J/K.

Why: The water and iron heat capacities set their relative contributions to each energy balance.

First balance: 2090(T_1 − 20) + 2.25(T_1 − 80) = 0, so T_1 = (2090 × 20 + 2.25 × 80)/2092.25 = 20.0645238°C.

Why: The first insulated mixing event warms water and cools iron to one common temperature.

When the iron is removed, the water remains at T_1. External reheating resets the iron to 80°C; it does not reset the water to 20°C.

Why: Reheating changes the iron’s starting state for the next event, while the retained water keeps its first equilibrium temperature.

Second balance: 2090(T_2 − T_1) + 2.25(T_2 − 80) = 0. Hence T_2 = (2090T_1 + 2.25 × 80)/2092.25 = 20.1289783°C.

Why: Write a fresh energy balance using T_1 for the water, rather than repeating the original 20°C input.

Equivalently, 80 − T_2 = (80 − 20)[C_w/(C_w + C_Fe)]². The second rise is slightly smaller than the first because the water is already warmer.

Why: Each mixing event shrinks the gap to 80°C by the same heat-capacity factor, so the second rise is slightly smaller.

Use the reference length at its stated reference temperature: L(T) = 20.0[1 + α(T − 20°C)] m.

Why: The given 20 m length belongs to 20°C, so every calculated expansion must use that reference state.

At −5°C, ΔT = −25 K, so L_cold = 20.0[1 − (11 × 10⁻⁶)(25)] = 19.99450 m.

Why: The cold antenna is 25 K below the reference, making the expansion term a contraction.

At 50°C, ΔT = 30 K, so L_hot = 20.0[1 + (11 × 10⁻⁶)(30)] = 20.00660 m.

Why: The hot antenna is 30 K above the reference, making its length slightly greater than 20 m.

The warming-day change is L_hot − L_cold = 20.0α(50 − (−5)) = 0.01210 m = 12.10 mm.

Why: Subtract the two endpoint lengths to find the full 55 K day-warming extension.

For equal numerical readings x, use x = (9/5)x + 32.

Why: Insert one unknown numerical reading into both sides of the Celsius-to-Fahrenheit conversion.

Solve −(4/5)x = 32, giving x = −40. This temperature is −40°C = −40°F = 233.15 K.

Why: Solve that linear equation, then add 273.15 to the Celsius value for its absolute kelvin temperature.

A temperature difference does not use the 32° offset: ΔT_K = ΔT_C = (5/9)ΔT_F = (5/9)(30) = 16.6667 K.

Why: A temperature rise cancels the scale offset, so only the 5/9 size conversion remains.

The two water samples have the same specific heat, so c cancels from the energy balance.

Why: Both samples are water with the same c, so that common factor divides out of their energy balance.

Write 250(T_f − 80) + 400(T_f − 15) = 0.

Why: The warmer sample loses energy while the cooler sample gains it; their signed changes must sum to zero.

Thus T_f = [250(80) + 400(15)]/(250 + 400) = 40.0°C.

Why: Solve for the common temperature as a mass-weighted mean because the specific heats are equal.

The brass heat capacity is C_b = 0.20(380) = 76 J/K. The water plus calorimeter heat capacity is C_c = 0.25(4180) + 150 = 1195 J/K.

Why: The water and cup start together, so add their heat capacities before balancing them against the hotter brass.

Energy balance is 76(T_f − 100) + 1195(T_f − 20) = 0.

Why: The insulated system has zero net energy change; brass cooling pays for warming both water and calorimeter.

The equilibrium is T_f = [76(100) + 1195(20)]/(76 + 1195) = 24.78364°C.

Why: Solve the balance for the common final temperature using both initial temperatures.

The calorimeter absorbs some energy, so omitting it would overestimate the final water temperature.

Why: Ignoring the cup would leave out an energy recipient and predict too much warming of the water.

The needed heat is Q = mcΔT = 0.50(4180)(95 − 18) = 160,930 J.

Why: Compute the energy needed for the 77 K water-temperature rise using its mass and specific heat.

Useful heating power is P_useful = 0.80(1000) = 800 W.

Why: Only 80% of the electrical power warms the water, so use 800 W for the heating rate.

Time is t = Q/P_useful = 160,930/800 = 201.1625 s.

Why: Divide the required energy by useful power to obtain elapsed heating time.

Initial translational energy is K = ½mv² = ½(75)(12²) = 5400 J.

Why: The prompt excludes wheel spin, so the available starting energy is the bicycle-rider translational kinetic energy.

The specified fraction absorbed by the brake is Q = 0.50K = 2700 J.

Why: The brake receives only the stated half of that energy, rather than the full stopping-energy loss.

Its temperature rise is ΔT = Q/(m_bc_b) = 2700/(1.0 × 490) = 5.51020 K.

Why: Divide the absorbed energy by the brake assembly’s heat capacity to find its temperature increase.

First warm the ice to 0°C: Q_1 = 0.100(2100)(20) = 4200 J.

Why: The initially subzero ice needs sensible heat to reach 0°C before melting begins.

The remaining energy is 20,000 − 4200 = 15,800 J. Complete melting would require 0.100(334,000) = 33,400 J, so melting is only partial.

Why: Compare the energy left after warming with the latent heat needed to melt the whole sample.

Melted mass is m_liquid = 15,800/334,000 = 0.0473054 kg.

Why: Divide the remaining heat by fusion energy per kilogram to get the mass that actually melts.

Remaining ice is 0.100 − 0.0473054 = 0.0526946 kg. Both phases coexist at 0°C.

Why: Subtract the melted mass from the original mass; coexisting ice and liquid stay at the melting temperature.

The warm water could release 200(4.18)(40) = 33,440 J by cooling to 0°C.

Why: Cooling the warm water to 0°C gives the maximum energy it can supply without freezing.

Warming and melting all the ice requires 50(2.10)(10) + 50(334) = 17,750 J, which is less than 33,440 J; all ice melts.

Why: Compare that available energy with both warming the ice and melting it to decide which final phase model applies.

The excess energy 33,440 − 17,750 = 15,690 J warms the final 250 g of water from 0°C.

Why: After all the ice melts, the unused energy warms the combined mass of liquid water.

Thus T_f = 15,690/[250(4.18)] = 15.01435°C. This positive value confirms complete melting.

Why: Divide that surplus by the final water heat capacity; a positive result confirms the all-liquid assumption.

The steam releases latent heat 10(2257) = 22,570 J on condensation.

Why: Condensation releases latent heat while the steam changes phase at 100°C.

The condensed water then cools from 100°C to T_f, releasing 10(4.18)(100 − T_f). The original water gains 200(4.18)(T_f − 20).

Why: The condensed steam also cools, and its released sensible heat joins the latent heat warming the original water.

Energy balance gives 22,570 + 41.8(100 − T_f) = 836(T_f − 20).

Why: Set the two steam-origin heat releases equal to the original water’s heat gain.

Solve: T_f = (22,570 + 4180 + 16,720)/877.8 = 49.52153°C, below boiling; the all-condensed assumption is consistent.

Why: Solve for the shared temperature and check it stays below 100°C, as the assumed complete condensation requires.

Energy on the rising segment is Q_rise = Pt = 50(40) = 2000 J.

Why: Constant heater power makes delivered energy equal to power times the rising-segment duration.

Use c = Q_rise/(mΔT) = 2000/(0.040 × 20) = 2500 J/(kg·K).

Why: The rising segment changes temperature, so its energy determines mc and hence the specific heat.

Energy during the plateau is Q_phase = 50(120) = 6000 J.

Why: During the plateau the heater still adds energy, now for the measured phase-change duration.

Latent heat is L = Q_phase/m = 6000/0.040 = 150,000 J/kg.

Why: The plateau energy changes phase without raising temperature, so divide it by mass to get latent heat.

Cooling the liquid to 60°C releases 30(2.4)(80 − 60) = 1440 J.

Why: First remove sensible heat from the liquid until it reaches its 60°C freezing point.

Solidifying at 60°C releases 30(200) = 6000 J.

Why: The phase change releases fusion energy at 60°C before the solid can cool further.

Cooling the solid to 20°C releases 30(2.0)(60 − 20) = 2400 J.

Why: Use the solid’s different specific heat for the final cooling stage below 60°C.

Total energy released is 9840 J. Taking Q positive into the wax gives Q = −9840 J.

Why: Add the three released amounts; heat leaving the wax is negative under the stated into-wax sign convention.

The temperature change is ΔT = 100 − 20 = 80 K.

Why: Expansion depends on the change from the wire’s 20°C reference state, here an 80 K rise.

Extension is ΔL = L_0αΔT = 0.750(19 × 10⁻⁶)(80) = 0.001140 m.

Why: Multiply reference length by α and the temperature rise to find the free extension.

Thus L_f = 0.751140 m. The extension is 1.140 mm.

Why: Add the extension to the reference length and convert it to millimeters for a useful size check.

Imagine filling the hole with the same material: every length of the uniformly heated plate expands with the same scale factor, including the imaginary plug’s diameter.

Why: The imaginary plug expands with the plate, showing that the hole’s diameter scales like any other unconstrained length.

The temperature rise is 150 K, so d_f = d_0(1 + αΔT) = 20.0[1 + (23 × 10⁻⁶)(150)] mm.

Why: Use the 150 K rise and linear expansion factor on the original hole diameter.

This gives d_f = 20.069 mm. The hole grows, because its boundary moves outward with the plate.

Why: The positive expansion term moves the boundary outward, so the calculated diameter is larger.

Both experience ΔT = 70 K.

Why: Both rails start and finish at the same temperatures, so they share one temperature rise.

Aluminum extends by 2.00(23 × 10⁻⁶)(70) = 0.00322 m; steel extends by 2.00(11 × 10⁻⁶)(70) = 0.00154 m.

Why: Use each rail’s own α to find how much its right end moves when its left reference end stays aligned.

The right-end separation is the difference, 0.00322 − 0.00154 = 0.00168 m.

Why: Subtract the steel extension from the larger aluminum extension to get the end-to-end mismatch.

The volume scale factor is V_f/V_i = 1 + βΔT = 1 + (1.10 × 10⁻³)(30) = 1.033.

Why: The ethanol warms by 30 K, giving its volume-growth factor in the supplied linear model.

Mass is conserved, so ρ_f = m/V_f = ρ_i/(1.033).

Why: No evaporation means the mass stays fixed, so density falls by the inverse of the volume-growth factor.

Thus ρ_f = 789/1.033 = 763.7948 kg/m³. Heating increases volume and lowers density in this model.

Why: Apply that inverse factor to the initial density; dividing rather than multiplying reflects the enlarged volume.

Free thermal strain would be ε_th = αΔT = (11 × 10⁻⁶)(40) = 4.40 × 10⁻⁴.

Why: Compute the positive strain the warmer steel would have if it were allowed to expand freely.

To keep total strain zero, mechanical strain must be ε_mech = −ε_th.

Why: The fixed ends force zero total strain, so elastic compression must cancel the positive thermal strain.

Hooke’s law gives σ = Eε_mech = −(200 × 10⁹)(4.40 × 10⁻⁴) = −88.0 × 10⁶ Pa.

Why: Young’s modulus turns that required negative mechanical strain into stress.

The negative sign under a tension-positive convention means compression. The result assumes elastic behavior and rigid restraints.

Why: Under the chosen sign convention the negative stress is compression, conditional on the rod and restraints remaining elastic and rigid.

Convert thickness to 0.0060 m and time to 1800 s. The face temperature difference is 20 K.

Why: Convert thickness and duration to SI units before using conductivity in watts and meters.

The conduction rate is P = kAΔT/L = 1.0(2.0)(20)/0.0060 = 6666.67 W.

Why: The maintained face-temperature difference drives a steady heat rate inversely proportional to glass thickness.

Energy in 1800 s is Q = Pt = 6666.67(1800) = 12,000,000 J.

Why: Multiply the constant heat rate by 1800 s to get the total transmitted energy.

These are maintained glass-face temperatures; using indoor and outdoor air temperatures would require additional surface resistances.

Why: The law uses glass-face temperatures; air-to-glass heat transfer would introduce extra resistance absent from this model.

Area-normalized resistances are R_brick = 0.12/0.60 = 0.20 m²·K/W and R_foam = 0.040/0.030 = 1.33333 m²·K/W.

Why: For each slab, thickness divided by conductivity measures its resistance to heat flow per unit area.

Series layers have the same heat flux: q = (20 − (−10))/(0.20 + 1.33333) = 19.5652 W/m².

Why: The same steady flux crosses both layers, so their resistances add across the total 30 K temperature drop.

The heat rate is P = Aq = 8.0(19.5652) = 156.522 W.

Why: Multiply heat flux by the wall area to convert watts per square meter into total watts.

The brick temperature drop is qR_brick = 3.91304 K, so T_interface = 20 − 3.91304 = 16.08696°C.

Why: Subtract the brick’s share of the temperature drop from the hot outer-face temperature to locate the interface temperature.

The remaining 26.08696 K drop occurs across the foam, consistent with its larger resistance.

Why: Check the foam takes the remaining drop; its larger thermal resistance should account for most of the total.

Net outward radiative power is P_out = εσA(T_body⁴ − T_surroundings⁴). All fourth-power temperatures must be in kelvins.

Why: Subtract absorbed enclosure radiation from emitted body radiation; absolute temperatures are required in the fourth powers.

P_out = (0.80)(5.670374419 × 10⁻⁸)(0.10)(400⁴ − 300⁴) = 79.38524 W.

Why: Insert the two kelvin temperatures, area and emissivity to get the initial outward power.

The body’s energy rate is mc(dT/dt) = −P_out, so dT/dt = −79.38524/(0.20 × 900) = −0.44103 K/s.

Why: The body loses energy, so its temperature-rate equation has the negative of the outward power.

This is the initial rate only; as the body cools, the fourth-power difference and cooling rate decrease.

Why: As the body approaches the enclosure temperature, the radiation imbalance shrinks; the first-instant rate cannot describe the whole cooling history.

The maintained surface-to-air temperature difference is 25 K.

Why: The surface stays 25 K warmer than the air, so that fixed difference drives the specified heat-loss law.

Use the supplied convection law: P_out = 8.0(1.5)(25) = 300 W.

Why: Use the supplied convection coefficient with exposed area and temperature difference to calculate power.

In 5.0 min = 300 s, Q_out = Pt = 300(300) = 90,000 J.

Why: Convert the maintained-loss interval to seconds and multiply it by the constant power.

Bulk air motion carries the energy away; h depends on the flow and surface, unlike a solid’s thermal conductivity k.

Why: The coefficient describes this moving-air heat transfer; it is not a replacement value for a material’s conduction k.

Heat capacities are C_cu = 0.50(390) = 195 J/K and C_w = 0.30(4180) = 1254 J/K.

Why: Compute each object’s heat capacity to weight its temperature change in the combined-system balance.

Energy entering the combined system is −5000 J, so 195(T_f − 90) + 1254(T_f − 20) = −5000.

Why: The room receives 5000 J, so the combined objects’ signed energy change is negative rather than zero.

Collect terms: 1449T_f = 195(90) + 1254(20) − 5000 = 37,630 J.

Why: Gather both final-temperature terms and subtract the known energy leak from the initial energy-weighted sum.

Thus T_f = 37,630/1449 = 25.96963°C. It is lower than the no-leak weighted mean, as expected.

Why: Divide by total heat capacity; losing energy to the room must lower the result from the insulated mixing temperature.