Zadanie 2c
Wzorowane na: Chemistry HL Paper 1B, May 2026, TZ2, pytanie 2(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 2a-2f.
Specific energy is the heat energy released per unit mass of fuel burnt. A student determined the specific energy of ethanol by burning it in a spirit burner under a copper can containing 120.0 g of water. The diagram shows the apparatus used.
The burner was weighed before and after each of five trials, and the temperature of the water was measured before and after heating. Fresh water was used for every trial.
| Spirit burner with ethanol | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Trial 5 |
|---|---|---|---|---|---|
| Initial mass / g (± 0.001 g) | 148.236 | 147.801 | 148.115 | 147.652 | 148.021 |
| Final mass / g (± 0.001 g) | 147.374 | 146.870 | 147.370 | 146.774 | 147.121 |
| Water in copper can | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Trial 5 |
|---|---|---|---|---|---|
| Initial temperature / °C (± 0.05 °C) | 20.4 | 19.8 | 22.0 | 21.2 | 21.3 |
| Final temperature / °C (± 0.05 °C) | 34.0 | 35.3 | 29.0 | 34.7 | 35.8 |
Polecenie
Calculate the specific energy of ethanol for trial 5, in J g−1. Use sections 1 and 2 of the data booklet.
mwater = 120.0 g
Sprawdź rozwiązanieUkryj rozwiązanieKlucz CKE i punktacja
Odpowiedź z klucza
«Q = mcΔT = 120.0 × 4.18 × 14.5 =» 7273 «J» ✔
«mfuel = 148.021 − 147.121 = 0.900 g, so specific energy = 7273/0.900 =» 8081 «J g−1»/8.08 × 103 ✔
Schemat punktowania
Award [2] for a correct final answer. Accept 8080 to 8100 because of rounding; accept an answer based on ECF from 2a. M1: Q from m, c and ΔT. M2: division by the mass of ethanol burnt, found from the table.