Zadanie 6b

(i) «Q = mcΔT = 100.00 g × 4.18 J g⁻¹ K⁻¹ × 30.0 K =» 12 540 J ✔

«$\Delta H_c = -\dfrac{12.54\ \mathrm{kJ}}{0.0250\ \mathrm{mol}} =$» −502 kJ mol⁻¹ ✔

(ii) Source of systematic error: heat loss «to the surroundings, the can and the air» ✔

Improvement: insulate the can / put a lid on the can / use a draught shield around the apparatus

OR use a calibrated (bomb) calorimeter ✔

(iii) «$\dfrac{0.5 + 0.5}{30.0} \times 100 =$» 3.3 % ✔

(iv) use a thermometer with smaller scale divisions «a smaller uncertainty»

OR use a smaller mass of water / burn more ethanol so that the temperature change is larger

OR use a more precise balance ✔

(v) «$\dfrac{|-1367 - (-502)|}{1367} \times 100 =$» 63.3 % ✔

Alternative with the given −400 kJ mol⁻¹: «(1367 − 400)/1367 × 100 =» 70.7 %

(i) [2]: award [1 max] for +502 kJ mol⁻¹; award [2] for the correct final answer with sign. (ii) [2]: M1 the source (heat loss), M2 an improvement that addresses it. (iii) [1]: accept 2.4 % if the uncertainties are combined as $\sqrt{0.5^2 + 0.5^2}/30.0 \times 100$. Do not award for 3.33 without rounding to two significant figures. (iv) [1]: do not accept more repetitions. (v) [1]: ECF from (i).