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Question 21M.2.SL.TZ1.3

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Date May 2021 Marks available [Maximum mark: 12] Reference code 21M.2.SL.TZ1.3
Level SL Paper 2 Time zone TZ1
Command term Calculate, Determine, Estimate, Explain, Show that, State Question number 3 Adapted from N/A
3.
[Maximum mark: 12]
21M.2.SL.TZ1.3

A mass of 1.0 kg of water is brought to its boiling point of 100 °C using an electric heater of power 1.6 kW.

(a.i)

The molar mass of water is 18 g mol−1. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

[2]

Markscheme

Ek = « 32(1.38×10-23)(373)» = 7.7×10-21 «J» 

v = «3×1.38×10-23×6.02×1023×3730.018» = 720 «m s−1» 

(a.ii)

State one assumption of the kinetic model of an ideal gas.

[1]

Markscheme

particles can be considered points «without dimensions»

no intermolecular forces/no forces between particles «except during collisions»

the volume of a particle is negligible compared to volume of gas

collisions between particles are elastic

time between particle collisions are greater than time of collision

no intermolecular PE/no PE between particles  

 

Accept reference to atoms/molecules for “particle”

A mass of 0.86 kg of water remains after it has boiled for 200 s.

(b.i)

Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer.

[2]

Markscheme

«mL = P t» so «L=1600×2000.14» = 2.3 x 106 «J kg-1» 

J kg−1 

(b.ii)

Explain why the temperature of water remains at 100 °C during this time.

[1]

Markscheme

«all» of the energy added is used to increase the «intermolecular» potential energy of the particles/break «intermolecular» bonds/OWTTE

Accept reference to atoms/molecules for “particle” 

(c)

The heater is removed and a mass of 0.30 kg of pasta at −10 °C is added to the boiling water.

Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible.

Specific heat capacity of pasta = 1.8 kJ kg−1 K−1
Specific heat capacity of water = 4.2 kJ kg−1 K−1

[3]

Markscheme

use of mcΔT 

0.86 × 4200 × (100 – T) = 0.3 × 1800 × (T +10)

Teq = 85.69«°C» ≅ 86«°C» 

Accept Teq in Kelvin (359 K).

The electric heater has two identical resistors connected in parallel.

The circuit transfers 1.6 kW when switch A only is closed. The external voltage is 220 V.

(d.i)

Show that each resistor has a resistance of about 30 Ω.

[1]

Markscheme

P=v2R so 22021600 so R=30.25 «Ω» 

Must see either the substituted values OR a value for R to at least three s.f.

 

(d.ii)

Calculate the power transferred by the heater when both switches are closed.

[2]

Markscheme

use of parallel resistors addition so Req = 15 «Ω»

P = 3200 «W»