IB Physics HL · guided topic map

Thermodynamics for IB Physics HL

Thermodynamics for IB Physics HL, organized into 1 syllabus topic and 3 mapped concept guides.

Syllabus topics: 1
Mapped concept guides: 3
Educational level: IB Diploma Physics Higher Level

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B.4

Thermodynamics

The particulate nature of matter

3 guides

01The first law and gas processesHL only
02Heat engines and the second lawHL only
03Entropy and irreversibilityHL only

Diagrams

Thermodynamics as IB Physics HL draws it

The figures from the IB Physics HL practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Figure 4Thermodynamics · Current and circuitsIB

Figure comment

Figure 4Circuit diagram of a single series loop. On the left a dashed boundary encloses the cell itself: a cell symbol with its long positive plate uppermost, labelled e.m.f. 6.0 V, in series with a resistor labelled r = 0.75 Ω. The two terminals cross the dashed boundary and the loop continues round to a resistor in the upper wire labelled R = 3.0 Ω. An arrow on the lower wire, labelled I, marks the direction of the conventional current.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

B.4 Thermodynamics B.5 Current and circuits Open IB question · The particulate nature of matter Q10

Guided questions 5 parts

Reading cue. The dashed boundary is not a component: r = 0.75 Ω sits inside the cell, so 6.0 V is the e.m.f. of the whole loop and never the reading across the terminals while current flows.

aState The loop is broken at R and a voltmeter of very high resistance is connected across the cell's terminals. State its reading, and give a reason.
recall2 marks
Check answer 2 marks
1. 6.0 V
2. with no current there is no potential difference across r, so the terminal p.d. equals the e.m.f.
bDetermine The current marked I is 1.6 A. Determine the charge that passes through R in 2.0 minutes, and the chemical energy the cell converts in that time.
routine3 marks
Check answer 3 marks
1. Q = It = 1.6 × 120 = 192 C
2. the e.m.f. is the energy the cell gives to each coulomb, 6.0 J C⁻¹
3. E = εQ = 6.0 × 192 = 1.2 × 10³ J
cDetermine A second 3.0 Ω resistor is connected in parallel with R. Determine the new terminal potential difference of the cell, and state what happens to the e.m.f.
demanding4 marks
Check answer 4 marks
1. external resistance becomes 1.5 Ω
2. I = 6.0 / (1.5 + 0.75) = 2.67 A
3. terminal p.d. = 2.67 × 1.5 = 4.0 V
4. the e.m.f. stays at 6.0 V; only the lost volts Ir has changed, from 1.2 V to 2.0 V
dShow (that) Show that the power delivered to the external resistor is greatest when R equals r, determine that maximum power for the cell drawn, and state the power actually delivered to the 3.0 Ω resistor.
top of the paper4 marks
Show a hint
No calculus is needed: try rewriting (R + r)²/R as (R − r)²/R + 4r.
Check answer 4 marks
1. P = I²R = ε²R/(R + r)², which can be written P = ε²/[(R + r)²/R]
2. (R + r)²/R = (R − r)²/R + 4r, so the denominator is least, and P greatest, when R = r
3. maximum power = ε²/(4r) = 6.0² / (4 × 0.75) = 12 W
4. with R = 3.0 Ω the circuit delivers only I²R = 1.6² × 3.0 = 7.7 W, because R is four times r

Transfer challenge

Four identical cells, each of e.m.f. 1.5 V and internal resistance 0.30 Ω, are connected in series with a lamp of resistance 2.0 Ω. Determine the current and the potential difference across the terminals of one cell.

Check answer 4 marks

e.m.f.s in series add to 4 × 1.5 = 6.0 V, and the internal resistances add to 4 × 0.30 = 1.2 Ω
I = 6.0 / (2.0 + 1.2) = 1.875 A, so 1.9 A to two significant figures
terminal p.d. of one cell = 1.5 − 1.875 × 0.30 = 0.94 V
the four terminal p.d.s sum to 3.75 V across the lamp, so 2.25 V of lost volts is now spread over four internal resistances instead of one