Thermal Expansion, Calorimetry and Heat Transfer: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A pendulum clock uses a simple pendulum whose bob hangs from a light metal rod of linear expansion coefficient α; treat the pendulum's effective length as proportional to the rod's length. The clock gains 6.0s per day when the room is at 10∘C and loses 12.0s per day when the room is at 30∘C. Take one day =86400s, and work to first order in αΔθ. The temperature at which the clock keeps correct time and the value of α are, respectively, ______. Exactly one option is correct.
A16.7∘C and 2.1×10−5K−1
B20.0∘C and 1.4×10−5K−1
C16.7∘C and 1.0×10−5K−1
D23.3∘C and 2.1×10−5K−1
Show answer and solution
Answer: A
Period T∝ℓ, so TΔT=21α(θ−θ0), where θ0 is the temperature of correct timekeeping. Time lost per day =86400⋅21α(θ−θ0) (negative means a gain).
At 10∘C: 43200α(θ0−10)=6. At 30∘C: 43200α(30−θ0)=12.
Dividing: 30−θ0θ0−10=21, so θ0=50/3≈16.7∘C.
Then α=43200×13.3312=2.08×10−5K−1≈2.1×10−5K−1.
Dropping the factor 21 gives 1.0×10−5 (option C); taking the midpoint 20∘C ignores the unequal gain and loss.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A solid glass sinker is completely immersed in a liquid, and its apparent loss of weight (the buoyant force) is measured. At 0∘C the loss of weight is 0.600N; at 50∘C it is 0.588N. The volume expansion coefficient of the glass is 2.4×10−5K−1. Model the sinker's volume as V0(1+γsΔθ) and the liquid's density as ρ0/(1+γlΔθ), with Δθ measured from 0∘C, and assume g is constant. The volume expansion coefficient γl of the liquid is closest to ______. Exactly one option is correct.
A4.08×10−4K−1
B4.33×10−4K−1
C3.84×10−4K−1
D4.57×10−4K−1
Show answer and solution
Answer: B
Buoyant force =Vρg. So B0B50=1+50γl1+50γs.
1+50γl=(1+50×2.4×10−5)0.5880.600=1.0012×1.020408=1.021633.
γl=500.021633=4.33×10−4K−1.
Ignoring the sinker's expansion gives 4.08×10−4; using the wrong sign for it gives 3.84×10−4.
Physical check: the sinker grows, which raises the buoyant force, yet the buoyant force falls. So the liquid's density must fall by more than 2%, and γl exceeds the uncorrected 4.08×10−4.
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A solid rod in the shape of a frustum of a cone has length L; its radius increases linearly from r at end A to 2r at end B. The rod is homogeneous, its curved surface is perfectly insulated, and end A is held at 100∘C and end B at 10∘C. Treat the steady heat flow as one-dimensional along the axis (each cross-section is at a single temperature). In the steady state, the temperature of the cross-section midway between A and B is ______. Exactly one option is correct.
A55∘C
B70∘C
C40∘C
D32.5∘C
Show answer and solution
Answer: C
In the steady state the heat current H is the same through every cross-section: H=−kA(x)dxdT, with A(x)=πr2(1+x/L)2, where x is measured from A.
So T(0)−T(x)=kπr2H∫0x(1+x′/L)2dx′=kπr2HL(1−1+x/L1).
At x=L the bracket is 1/2; at x=L/2 it is 1/3. So 100−10100−Tm=1/21/3=32.
Tm=100−60=40∘C.
The narrow half has the larger thermal resistance, so two-thirds of the temperature drop occurs across it. The linear-profile answer, 55∘C, is wrong.
Difficulty: Very hard
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