21 Marking scheme: Worksheet (A2)

1 a The atoms in a solid are arranged in a three-dimensional structure. [1]

There are strong attractive forces between the atoms. [1]
The atoms vibrate about their equilibrium positions. [1]
b The atoms in a liquid are more disordered than those in a solid. [1]
There are still attractive electrical forces between molecules but these are weaker than
those between similar atoms in a solid. [1]
The atoms in a liquid are free to move around. [1]
c The atoms in a gas move around randomly. [1]
There are virtually no forces between the molecules (except during collisions) because they
are much further apart than similar molecules in a liquid. [1]
The atoms of a gas move at high speeds (but no faster than those in a liquid at the same
temperature). [1]
2 The atoms move faster [1]
because their mean kinetic energy increases as the temperature is increased. [1]
The atoms still have a random motion. [1]
3 a The internal energy of a substance is the sum (of the random distribution) of the kinetic
and potential energies of its particles (atoms or molecules). [1]
b There is an increase in the average kinetic energy of the aluminium atoms as they vibrate
with larger amplitudes about their equilibrium positions. [1]
The potential energy remains the same because the mean separation between the atoms
does not change significantly. [1]
Hence, the internal energy increases because there is an increase in the kinetic energy of
the atoms. [1]
c As the metal melts, the mean separation between the atoms increases. [1]
Hence, the electrical potential energy of the atoms increases. [1]
There is no change in the kinetic energy of the atoms because the temperature remains
the same. [1]
The internal energy of the metal increases because there is an increase in the electrical
potential energy of the atoms. [1]
4 Change in thermal energy = mass × specific heat capacity × change in temperature [1]
5 The specific heat capacity refers to the energy required to change the temperature of a substance. [1]
Specific latent heat of fusion is the energy required to melt a substance; there is no change in
temperature as the substance melts. [1]
6 E = mc∆θ [1]
E = 6.0 × 105 × 4200 × (24 – 21) [1]
E = 7.56 × 109 J ≈ 7.6 × 109 J [1]
7 E = mc∆θ [1]
E = 300 × 10–3 × 490 × (20 – 300) [1]
E = –4.1 × 104 J (The minus sign implies energy is released by the cooling metal.) [1]
8 E = mLf = 200 × 10−3 ×3.4 × 105 [1]
= 6.8 × 104 J [1]

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 21 Marking scheme: Worksheet (A2)

9 a i T = 273 + 0 = 273 K [1]

ii T = 273 + 80 = 353 K [1]
iii T = 273 – 120 = 153 K [1]
b i θ = 400 – 273 = 127 °C [1]
ii θ = 272 – 273 = –1 °C [1]
iii θ = 3 – 273 = –270 °C [1]
10 a The thermal energy E supplied and the specific heat capacity c remain constant.
The mass m is larger by a factor of 3. [1]
Since E = mc∆θ, we have:
E 1
∆θ = ; ∆θ ∝ [1]
mc m
15
Therefore ∆θ = = 5.0 °C [1]
3
b The thermal energy E supplied is halved but the specific heat capacity c and the mass m
remain constant. [1]
Since E = mc∆θ, we have:
E
∆θ = ; ∆θ ∝ E [1]
mc
15
Therefore ∆θ = = 7.5 °C [1]
2
11 a Melting point = 600 °C [1]
(There is no change in temperature during change of state.)
b The lead is being heated at a steady rate and therefore the temperature also increases
at a steady rate. [1]
c The energy supplied to the lead is used to break the atomic bonds and increase the
separation between the atoms of lead (and hence their potential energy increases). [1]
d E = mc∆θ [1]
E = 200 × 10–3 × 130 × (600 – 0) [1]
E = 1.56 × 104 J ≈ 1.6 × 104 J [1]
e In a time of 300 s, 1.56 × 104 J of energy is supplied to the lead.
Rate of heating = power
1.56 × 10 4
power = [1]
300
power = 52 W [1]
f Energy supplied = 52 × 100 = 5200 J [1]
∆E 5200
Lf = = [1]
∆m 0.2
= 26 000 J kg−1 [1]
12 The energy supplied per second is equal to the power of the heater.
In a time of 1 s, water of mass 0.015 kg has its temperature changed from 15 °C to 42 °C. [1]
E = mc∆θ (where E is the energy supplied in 1 s) [1]
E = 0.015 × 4200 × (42 – 15) [1]
E = 1.7 × 103 J [1]
The power of the heater is therefore 1.7 kW. [1]
m
(You may use P = ( )c∆θ )
t
13 The gas does work against atmospheric pressure. [1]
Energy to do this work is taken from the internal energy of the gas. [1]

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 21 Marking scheme: Worksheet (A2)

14 Heat ‘lost’ by hot water = heat ‘gained’ by cold water. [1]

0.3 × c × (90 – θ) = 0.2 × c × (θ – 10) [1]
where c is the specific heat capacity of the water and θ is the final temperature.
The actual value of c is not required, since it cancels on both sides of the equation.
Hence:
0.3 × (90 – θ) = 0.2 × (θ – 10) [1]
27 – 0.3θ = 0.2θ – 2.0 [1]
0.5θ = 29 so θ = 58 °C [1]
15 Heat ‘lost’ by metal = heat ‘gained’ by cold water [1]
0.075 × 500 × (θ – 48) = 0.2 × 4200 × (48 – 18) [1]
(θ is the initial temperature of the metal.)
0.2 × 4200 × 30
θ – 48 = [1]
0.075 × 500
θ – 48 = 672 [1]
θ = 720 °C [1]

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