B: Thermal physics, greenhouse effect, gases, thermodynamics and circuits – IB Physics exam questions with mark scheme

1Core4 marks

An electric kettle of power $2.2\ \mathrm{kW}$ heats $1.5\ \mathrm{kg}$ of water from $20^\circ$C to $100^\circ$C. The specific heat capacity of water is $4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$.

(a) Calculate the thermal energy needed to raise the water to $100^\circ$C. [2]

(b) The kettle transfers 85% of its electrical energy to the water. Determine the time taken. [2]

$Q=mc\Delta T = 1.5\times4180\times80$(1)
$Q=5.02\times10^5$ J ($5.0\times10^5$ J)(1)
Electrical energy needed $=Q/0.85$ so $t = Q/(0.85P)$(1)
$t=268$ s, i.e. 270 s to 2 s.f. (allow 268 s; allow 267 s if $5.0\times10^5$ J is used)(1)
Answer: (a) $5.0\times10^5$ J (b) 270 s
Examiner tip: A temperature change of 80 K is the same as 80°C; do not convert it incorrectly.

2Hard7 marks

The solar constant at the Earth is $1.36\times10^3\ \mathrm{W\,m^{-2}}$ and the albedo of the Earth is 0.30. Assume the Earth has no atmosphere and that its surface radiates as a black body (emissivity 1). $\sigma=5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$.

(a) Show that the average intensity absorbed per unit area of the Earth's surface is about $240\ \mathrm{W\,m^{-2}}$. [2]

(b) Calculate the equilibrium surface temperature of the Earth under these assumptions. [2]

Only 70% of the incident radiation is absorbed: $1360\times(1-0.30)$ (= 952 W m$^{-2}$) (allow the two steps in either order, e.g. $1360/4=340$ then $\times0.70$)(1)
Radiation is intercepted by a disc of area $\pi R^2$ but absorbed over a surface of area $4\pi R^2$, so divide by 4: $952/4 = 238\ \mathrm{W\,m^{-2}}$ (must see 238 or more s.f.; 240 alone earns no credit)(1)
At equilibrium, power radiated = power absorbed: $\sigma T^4 = 238$ (allow 240)(1)
$T=(238/5.67\times10^{-8})^{1/4} = 255$ K (allow 254–256 K)(1)
Greenhouse gases (e.g. CO₂, CH₄, H₂O, N₂O) in the atmosphere absorb infrared radiation emitted by the surface (allow: IR frequency matches natural frequency of molecular vibration / resonance; reject: answers about the ozone layer, ozone hole or UV; reject 'reflect' / 'trap heat' without absorption of IR)(1)
They re-emit infrared in all directions, so some returns to the surface (OWTTE)(1)
The surface therefore receives more power than from the Sun alone, so it warms to a higher temperature before energy balance is reached (allow: equilibrium reached at higher T)(1)
Answer: (a) 238 W m$^{-2}$ (b) 255 K (c) greenhouse gases absorb and re-radiate IR back towards the surface
Examiner tip: In (c) the mark needs gases absorbing infrared emitted by the Earth's surface: 'trapping heat', 'reflecting' or the ozone layer/UV earn nothing.

3Standard5 marks

The spectrum of a star, assumed to be a black body, has a peak intensity at a wavelength of $4.8\times10^{-7}\ \mathrm{m}$. The radius of the star is $7.0\times10^{8}\ \mathrm{m}$. Wien's constant is $2.90\times10^{-3}\ \mathrm{m\,K}$.

(a) Determine the surface temperature of the star. [2]

(b) Calculate the luminosity of the star. [2]

(c) A second star has the same surface temperature but twice the radius. State the ratio of its luminosity to that of the first star. [1]

$T=2.90\times10^{-3}/\lambda_{max}$ (Wien's law)(1)
$T=6042$ K ($6.0\times10^3$ K)(1)
$L=\sigma A T^4$ with $A=4\pi R^2$(1)
$L=\sigma\times4\pi R^2T^4=4.7\times10^{26}$ W (allow $4.5\times10^{26}$ W if $T=6.0\times10^3$ K is used)(1)
Ratio = 4 (luminosity ∝ $R^2$)(1)
Answer: (a) $6.0\times10^3$ K (b) $4.7\times10^{26}$ W (c) 4
Examiner tip: Use the surface area of a sphere, $4\pi R^2$, and temperature in kelvin to the fourth power.

4Standard6 marks

A fixed mass of an ideal gas occupies a rigid container of volume $0.0250\ \mathrm{m^3}$ at a pressure of $3.0\times10^5\ \mathrm{Pa}$ and a temperature of $300\ \mathrm{K}$. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$, $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$.

(a) Calculate the amount of gas in moles. [2]

(b) Determine the number of molecules in the container. [1]

(d) Calculate the average translational kinetic energy of a molecule at $300\ \mathrm{K}$. [1]

$n=pV/RT = (3.0\times10^5\times0.0250)/(8.31\times300)$(1)
$n=3.01$ mol (3.0)(1)
$N=nN_A = 1.81\times10^{24}$ ($1.8\times10^{24}$)(1)
Constant volume and amount so $p/T$ is constant: $p_2=p_1T_2/T_1$(1)
$p_2=4.0\times10^5$ Pa ($4.0\times10^5$ Pa)(1)
$\bar E_K=\tfrac32 k_BT = 6.21\times10^{-21}$ J ($6.2\times10^{-21}$ J)(1)
Answer: (a) 3.0 mol (b) $1.8\times10^{24}$ (c) $4.0\times10^5$ Pa (d) $6.2\times10^{-21}$ J
Examiner tip: Temperatures in gas-law equations must be in kelvin.

5HardHL only8 marks

0.10 mol of a monatomic ideal gas undergoes the cycle ABCA shown on the $p$–$V$ diagram. AB is at constant pressure, BC is at constant volume and CA is a straight line. $R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

(a) Calculate the work done by the gas during AB. [1]

(b) Show that the change in internal energy of the gas during AB is 600 J. [2]

(d) Determine the net work done by the gas in one cycle. [2]

(e) Thermal energy is transferred into the gas only during AB. Calculate the efficiency of the cycle. [2]

$W=p\Delta V = 2.0\times10^5\times(3.0-1.0)\times10^{-3} = 400$ J (400 J)(1)
$\Delta U=\tfrac32 nR\Delta T$ with $T_A=p_AV_A/nR=241$ K and $T_B=3T_A=722$ K (or $\Delta U=\tfrac32\Delta(pV)=\tfrac32\times(6.0\times10^2-2.0\times10^2)$ J)(1)
$\Delta U=\tfrac32\times0.10\times8.31\times481 = 600$ J (600 J)(1)
$Q=\Delta U+W = 600+400 = 1000$ J (allow ECF from (a))(1)
Net work = area enclosed by the cycle (triangle $\tfrac12\times\Delta V\times\Delta p$) (allow summing the work for each stage: $W_{AB}+W_{BC}+W_{CA}=400+0-300$ J)(1)
$W_{net}=\tfrac12\times(2.0\times10^{-3})\times(1.0\times10^5) = 100$ J (100 J, cycle is clockwise so positive)(1)
Efficiency = net work / thermal energy input(1)
$=100/1000 = 0.10$ (10%) (allow ECF from (c) and (d))(1)
Answer: (a) 400 J (b) 600 J (c) 1000 J (d) 100 J (e) 10%
Examiner tip: Net work in a cycle is the enclosed area: check the direction of the cycle for the sign.

6Standard6 marks

A cell of emf 12 V and internal resistance $0.50\ \Omega$ is connected to the circuit shown. A $3.0\ \Omega$ resistor is in series with a parallel pair of $4.0\ \Omega$ and $6.0\ \Omega$ resistors.

(a) Show that the combined resistance of the parallel pair is $2.4\ \Omega$. [2]

(b) Calculate the current supplied by the cell. [2]

$1/R=1/4.0+1/6.0$ (= 0.4167) (allow product over sum $4.0\times6.0/(4.0+6.0)$)(1)
$R = 1/0.41\dot 6 = 24/10 = 2.4\ \Omega$ (result must follow from shown working; restating 2.4 Ω alone earns no credit)(1)
Total resistance $=0.50+3.0+2.4 = 5.9\ \Omega$ (internal resistance included)(1)
$I=\varepsilon/R_{tot} = 12/5.9 = 2.03$ A (2.0 A)(1)
p.d. across parallel pair $=IR = 2.03\times2.4=4.88$ V (4.9 V)(1)
$I=V/R = 4.88/6.0 = 0.81$ A (allow current-divider $2.03\times4.0/10.0$; allow 0.80 A if rounded 2.0 A and 4.8 V are used; allow ECF from (b))(1)
Answer: (a) 2.4 Ω (b) 2.0 A (c) 0.81 A
Examiner tip: Do not forget the cell's internal resistance when finding the total resistance.

7Core5 marks

A copper wire has length 2.5 m and diameter 0.40 mm. The resistivity of copper is $1.7\times10^{-8}\ \Omega\,\mathrm{m}$.

(a) Calculate the cross-sectional area of the wire. [1]

(b) Calculate the resistance of the wire. [2]

(d) A second wire has the same length and material but twice the diameter. State its resistance. [1]

$A=\pi d^2/4 = 1.26\times10^{-7}\ \mathrm{m^2}$ ($1.3\times10^{-7}$)(1)
$R=\rho L/A = 1.7\times10^{-8}\times2.5/1.26\times10^{-7}$(1)
$R=0.34\ \Omega$ (0.34)(1)
$P=I^2R = 3.0$ W (3.0 W; allow 3.1 W)(1)
Area is 4 times larger so $R = 0.34/4 = 0.085\ \Omega$ (0.085 Ω)(1)
Answer: (a) $1.3\times10^{-7}\ \mathrm{m^2}$ (b) 0.34 Ω (c) 3.0 W (d) 0.085 Ω
Examiner tip: Convert diameter in mm to metres and halve it to get the radius (or use $d^2/4$).

8Hard6 marks

A student measures the terminal potential difference $V$ of a cell for different currents $I$. The straight-line graph is shown. The line passes through $(0,\ 1.50\ \mathrm{V})$ and $(2.0\ \mathrm{A},\ 0.30\ \mathrm{V})$.

(a) State the emf of the cell. [1]

(b) Determine the internal resistance of the cell. [2]

(d) The cell is connected to a variable resistor. The power transferred to the resistor is a maximum when its resistance equals the internal resistance. Calculate this maximum power. [2]

emf = vertical intercept = 1.50 V(1)
$V=\varepsilon-Ir$ so $r=-$gradient (magnitude of gradient)(1)
$r=(1.50-0.30)/2.0 = 0.60\ \Omega$ (0.60 Ω)(1)
$I=\varepsilon/r = 1.50/0.60 = 2.5$ A (2.5 A)(1)
At maximum power, $R=r$ so $I=\varepsilon/2r = 1.25$ A; $P=I^2R$ (or $\varepsilon^2/4r$)(1)
$P=0.94$ W (0.94 W)(1)
Answer: (a) 1.50 V (b) 0.60 Ω (c) 2.5 A (d) 0.94 W
Examiner tip: The internal resistance is the magnitude of the gradient; give it without a negative sign.

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