Two absolute temperature scales A and B assign water's triple point values 200 A and 350 B. Find the relation between TA and TB.
Steps only
Both scales start at absolute zero, so their readings are proportional to Kelvin temperature.
Answer
TA=(4/7)TB.
Question 3
A resistance thermometer follows R=R₀[1+α(T−T₀)]. It reads 101.6 Ω at 273.16 K and 165.5 Ω at 600.5 K. What temperature corresponds to 123.4 Ω?
Steps only
Assume resistance varies linearly between the two calibration points.
Answer
About 385 K.
Question 4
(a) Why is water's triple point a better fixed point than ice and steam points? (b) What is the other fixed point on the Kelvin scale? (c) Why does Celsius conversion use 273.15, not 273.16? (d) What is the triple-point temperature on an absolute scale with Fahrenheit-sized units?
Steps only
(a) The triple point is a single, reproducible condition where ice, water and vapour coexist. Melting and boiling points vary with pressure and impurities.
Answer
(a) It is uniquely reproducible. (b) Absolute zero. (c) 273.15 K is 0°C; the triple point is 0.01°C. (d) About 491.7 degrees on the stated scale.
Question 5
Gas thermometers A and B show pressures 1.250×10⁵ Pa and 0.200×10⁵ Pa at water's triple point; at sulfur's melting point they show 1.797×10⁵ Pa and 0.287×10⁵ Pa. Estimate sulfur's melting temperature using each thermometer. Why do answers differ, and how can the difference be reduced?
Steps only
For a constant-volume gas thermometer, T/Ttp=P/Ptp, with Ttp=273.16 K.
Answer
Thermometer A: about 392.7 K; B: about 392.1 K. Use lower pressures and extrapolate toward zero pressure.
Question 6
A 1 m steel tape is correct at 27°C. On a 45°C day it measures a steel rod as 63.0 cm. Find the rod's actual length that day and its length at 27°C. αsteel=1.20×10⁻⁵ K⁻¹.
Steps only
The tape's 1 m division expands by factor 1+αΔT=1+(1.20×10⁻⁵)(18)=1.000216.
Answer
At 45°C: about 63.014 cm. At 27°C: 63.0 cm.
Question 7
A steel wheel hole of diameter 8.69 cm must fit over a steel shaft of diameter 8.70 cm. Both are at 27°C. To what temperature should the shaft be cooled? αsteel=1.20×10⁻⁵ K⁻¹.
Steps only
The shaft diameter must shrink by 0.01 cm. Use ΔD=αDΔT.
Answer
Cool the shaft to about −69°C.
Question 8
A hole in a copper sheet has diameter 4.24 cm at 27°C. Find the change in its diameter when heated to 227°C. αCu=1.70×10⁻⁵ K⁻¹.
Steps only
A hole expands like the copper around it.
Answer
The diameter increases by about 0.014 cm (0.14 mm).
Question 9
A brass wire 1.8 m long at 27°C is fixed taut between supports and cooled to −39°C. Its diameter is 2.0 mm; α=2.0×10⁻⁵ K⁻¹, Y=0.91×10¹¹ Pa. Find the tension.
Steps only
Because the ends cannot move, the prevented thermal strain is ε=α|ΔT|=(2.0×10⁻⁵)(66)=1.32×10⁻³.
Answer
About 3.8×10² N.
Question 10
A brass rod and steel rod, each 50 cm long and 3.0 mm diameter, are joined end to end. They warm from 40°C to 250°C. Find the total length change and whether thermal stress develops at the join. αbrass=2.0×10⁻⁵ K⁻¹, αsteel=1.2×10⁻⁵ K⁻¹.
Steps only
Temperature rise is 210 K. Brass extension=(0.50)(2.0×10⁻⁵)(210)=0.00210 m.
Answer
Total increase=3.36 mm; no thermal stress at the join.
Question 11
Glycerine has volume expansion coefficient 49×10⁻⁵ K⁻¹. Find its fractional density change for a 30°C rise.
Steps only
For a fixed mass, density is inversely proportional to volume.
Answer
Density falls by about 1.45% (approximately 1.47% for a small change).
Question 12
A 10 kW drill heats an 8.0 kg aluminium block for 2.5 min. Half the power is lost or heats the machine. Find the block's temperature rise. cAl=0.91 J g⁻¹ K⁻¹.
Steps only
Useful energy Q=0.50×10,000×150=750,000 J.
Answer
The block warms by about 103°C.
Question 13
A 2.5 kg copper block at 500°C is placed on a large ice block. What maximum mass of ice can melt? cCu=0.39 J g⁻¹ K⁻¹; latent heat of fusion=335 J/g.
Steps only
The copper cools to 0°C, releasing Q=mcΔT=(2500)(0.39)(500)=487,500 J.
Answer
About 1.46 kg of ice.
Question 14
A 0.20 kg metal block at 150°C is put into a copper calorimeter (water equivalent 0.025 kg) containing 150 cm³ water at 27°C. Final temperature is 40°C. Find the metal's specific heat. If heat escapes, is the result high or low?
Steps only
Water plus calorimeter equivalent mass is 0.150+0.025=0.175 kg.
Answer
c≈4.3×10³ J kg⁻¹ K⁻¹; with heat loss, this is an underestimate.
Question 15
Several diatomic gases have Cv near 5 cal mol⁻¹ K⁻¹, while a monatomic gas has about 2.92 cal mol⁻¹ K⁻¹. Explain the difference and the larger value for chlorine (6.17).
Steps only
A diatomic molecule can store energy in translation and rotation, giving a larger Cv than a monatomic gas, which mainly stores translational energy.
Answer
Diatomic molecules have rotational energy modes as well as translational modes. Chlorine's higher Cv suggests some vibrational mode contribution.
Question 16
A 30 kg child cools from 101°F to 98°F in 20 min because extra sweat evaporates. Estimate the average extra evaporation rate. Take body specific heat as water's 1 cal g⁻¹ °C⁻¹ and latent heat 580 cal/g.
Steps only
Temperature drop=3°F×5/9=1.667°C. Heat lost=30,000×1×1.667≈50,000 cal.
Answer
About 4.3 g of extra sweat per minute.
Question 17
A thermacole icebox is a 30 cm cube with 5 cm thick walls. It contains 4.0 kg ice. Outside is 45°C; k=0.01 W m⁻¹ K⁻¹. Estimate ice remaining after 6 h. Lf=335 kJ/kg.
Steps only
Take the ice interior at 0°C and heat flow through six flat walls. Approximate outer side as 0.30 m and inner side as 0.20 m; area for conduction is about 6(0.25²)=0.375 m².
Answer
About 3.8 kg of ice remains (using the mean wall area approximation).
Question 18
A brass boiler base has area 0.15 m² and thickness 1.0 cm. It boils water at 6.0 kg/min. Estimate the flame-side temperature. kbrass=109 W m⁻¹ K⁻¹, Lv=2256 kJ/kg.
Steps only
Heat used per second to vaporise water: H=(6/60)×2.256×10⁶=225,600 W.
Answer
Flame-side contact temperature is about 238°C (rough estimate).
Question 19
Explain: (a) a highly reflective body is a poor emitter; (b) brass feels colder than wood on a chilly day; (c) an ideal-blackbody pyrometer reads low for red-hot iron in open air but correctly in a furnace; (d) Earth without atmosphere would be very cold; (e) why steam heating is effective.
Steps only
(a) Good absorbers are good emitters; reflection means less radiation is absorbed and emitted.
Answer
(a) Reflectors emit poorly. (b) Brass removes heat from the hand faster. (c) Emissivity and furnace radiation explain the readings. (d) The atmosphere reduces infrared heat loss. (e) Condensing steam releases latent heat.
Question 20
A body cools from 80°C to 50°C in 5 min. How long to cool from 60°C to 30°C if surroundings are at 20°C?
Steps only
Using Newton's law approximation, cooling rate is proportional to the excess temperature above the surroundings.