Q21.
(a) M1 (q = mcΔT = 100 × 4.18 × 38(.0))
= 15 884 / 15 880 / 15 900 / 16 000 (J)
(OR 15.884 / 15.88 / 15.9 / 16 (kJ))
Award full marks for correct answer
Mark is for value not expression (at least 2sf); penalise incorrect units here only if M1 is the only potential scoring point in M1-M3
1
M2 Moles (methanol = 1.65 / 32.0) = 0.0516 or 0.052
At least 2sf
1
M3 Heat change per moles = M1/M2
(15 884 / 0.0516 / 1000 = 308 (kJ mol−1)
(allow 305 to 310)
At least 2sf; answer must be in kJ mol−1
1
M4 Answer = −308 (kJ mol−1) (allow −305 to −310)
This mark is for – sign (mark independently)
1
(b) Heating up copper / calorimeter / container / thermometer /
heat capacity of copper / calorimeter / thermometer not taken into account
OR
Evaporation of alcohol/methanol
OR
Experiment not done under standard conditions
Not human errors (e.g. misreading scales)
Not impure methanol
Allow evaporation of water
1
(c) (100 × 0.5 / 38 =) 1.3 or 1.32 or 1.316% (minimum 2 sf)
Allow correct answer to at least 2sf;
Allow 1.31 or 1.315%
1
(d) Idea that heat loss is more significant issue OR
Idea that temperature change/rise is (significantly / much)
bigger than uncertainty
One of these two ideas only and each one must involve a comparison
1
(e) M1 Mass of ethanol = 500 × 0.789 (= 394.5 or 395 (g))
1
M2 Moles of ethanol = M1 / 46.0 (= 8.576 or 8.58)
1
M3 Heat released = M2 × 1371 = 11800 (kJ) must be 3 sf
1
Correct answer to 3sf scores 3; correct value to 2sf or more than 3sf scores 2
Answers that are a factor of 10x out score 2 if given to 3sf or 1 if given to a different number of sf
M3 ignore units, but penalise incorrect units
M3 ignore sign
M2 and M3 – allow consequential marking
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