Question 22N.2.HL.TZ0.2a.ii
| Date | November 2022 | Marks available | [Maximum mark: 3] | Reference code | 22N.2.HL.TZ0.2a.ii |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Estimate | Question number | a.ii | 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
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.
