Question 22N.2.HL.TZ0.2
| Date | November 2022 | Marks available | [Maximum mark: 11] | Reference code | 22N.2.HL.TZ0.2 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Determine, Estimate, Explain, Outline, State | Question number | 2 | Adapted from | N/A |
A solar heating panel is placed on the roof of a house in order to heat water in a storage tank. The rest of the roof is covered with tiles.
On a certain day, the intensity of the solar radiation that is incident perpendicular to the surface of the panel is 680 W m−2.
The following data are available.
Mass of the water in the tank = 250 kg
Initial temperature of the water in the tank = 15 °C
Specific heat capacity of water = 4200 J kg−1 K−1
Overall efficiency of the heating system = 0.30
Albedo of the roof tiles = 0.20
Emissivity of the roof tiles = 0.97
Determine the minimum area of the solar heating panel required to increase the temperature of all the water in the tank to 30°C during a time of 1.0 hour.
[3]
energy required = 250 × 4200 × (30 − 15) ✓
energy available = 0.30 × 680 × t × A ✓
A = «» 21 «m2» OR 22 «m2» ✓
Allow ECF from MP1 and MP2.
Accept the correct use of 0.30 in either MP1 or MP2.
Most candidates had a good attempt at this but there were often slight slips. Some missed the efficiency of the process. Some included the albedo of the roof tiles. Some thought that the temperature rise needed to have 273 added to convert to kelvin. However, sometimes scoring through ECF (error carried forward), the average mark was around 2 marks.
Estimate, in °C, the temperature of the roof tiles.
[3]
absorbed intensity = (1 − 0.2) × 680 «= 544» «W m−2»
OR
emitted intensity = 0.97 × 5.67 × 10−8 × T4 ✓
T «K» ✓
42 «°C» ✓
Allow ECF from MP1 and MP2.
Allow MP1 if absorbed or emitted intensity is multiplied by area.
This was a bit more hit and miss than the previous question part. One common mistake was not understanding what albedo meant. Some took it as the amount of energy absorbed rather than reflected. Emissivity was often missed. Several candidates, successfully answering the question or not, were able to score MP3 converting the final temperature into Celsius degrees.
There is an air space above the water in the storage tank with an opening to the atmosphere. Assume that air behaves like an ideal gas.
State one way in which a real gas differs from an ideal gas.
[1]
can be liquefied ✓
has intermolecular forces / potential energy ✓
has atoms/molecules that are not point objects / take up volume ✓
does not follow the ideal gas law «for all T and p» ✓
collisions between particles are non-elastic ✓
Accept the converse argument.
This was very well answered. Candidates showed an understanding of the differences between ideal and real gases.
The air space is always at constant atmospheric pressure and constant volume, as the water level is kept constant. The air-space temperature and water temperature are the same.
The water is heated. Explain why the quantity of air in the storage tank decreases.
[2]
ALTERNATIVE 1
«constant p and V imply» nT = const ✓
T increases hence n decreases ✓
ALTERNATIVE 2
«constant p and n imply» V is proportional to T / air expands as it is heated ✓
«original» air occupies a greater volume OR some air leaves through opening ✓
MP2 in ALT 2 must come from expansion of air, not from expansion of water.
Award [0] for an answer based on expansion of water.
Award [1] max for an answer based on convection currents.
It was surprising to see a large number of answers based on the expansion of water, as the stem of the question clearly states that the level of water remains constant. Most successful candidates scored by quoting pV constant so concluding with the inverse relationship of n and T, others also managed to score by explaining that the volume of air increases and therefore must go out through the opening. Answers based on convection currents were given partial credit.
Another method of harnessing solar energy involves the use of photovoltaic cells.
Outline one advantage of the output of a photovoltaic cell compared to the output of a solar heating panel.
[2]
photovoltaic cells output electrical energy ✓
electrical is a more versatile form of energy ✓
Accept any reasonable advantage arising from the electrical output, .e.g., PV cells allow for the use of a long list of appliances, PV owners can sell excess power back to grid.
Accept electrical energy can be stored.
Do not accept references to efficiency for MP2.
This was well answered with some problems experienced to successfully identify an advantage.