Zadanie 2a
Wzorowane na: Chemistry HL Paper 1B, May 2026, TZ2, pytanie 2(a). 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 change in temperature for trial 5 and its uncertainty.
Give the temperature change to a suitable number of decimal places and the uncertainty to a suitable number of significant figures.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz CKE i punktacja
Odpowiedź z klucza
temperature change «35.8 − 21.3 =» 14.5 «°C» ✔
uncertainty «0.05 + 0.05 = ±» 0.1 «°C» ✔
OR
« = ±» 0.07 «°C» ✔
Schemat punktowania
M1: 14.5 °C with one decimal place; accept 14.50. M2: ±0.1 °C (one significant figure) or ±0.07 °C from combining the two readings in quadrature.