These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Experimental data and substances named with letters are fictional. Data: specific heat capacities (J g⁻¹ °C⁻¹) water and dilute solutions 4.18, ethanol 2.44, aluminium 0.897, iron 0.449, copper 0.385, silver 0.233; Ag 107.87 g mol⁻¹. Assume calorimeters are perfectly insulated unless told otherwise.
Question 1 (multiple choice · foundation)
How much energy is needed to warm 40.0 g of ethanol from 20.0 °C to 35.0 °C?
- (A) 1.46 kJ
- (B) 2.51 kJ
- (C) 3.42 kJ
- (D) 5.37 kJ
Answer and explanation
Answer: (A). ΔT = 35.0 − 20.0 = 15.0 °C, so q = mcΔT = 40.0 × 2.44 × 15.0 = 1464 J = 1.46 kJ.
- (B) uses water’s specific heat capacity (4.18) instead of ethanol’s.
- (C) uses the final temperature (35.0 °C) as ΔT.
- (D) adds the two temperatures (55.0 °C) instead of subtracting.
Question 2 (multiple choice · core)
50.0 g samples of water, aluminium, iron and copper, all at 20.0 °C, each absorb 500. J of energy. None changes phase. Which sample reaches the highest temperature?
- (A) water
- (B) aluminium
- (C) iron
- (D) copper
Answer and explanation
Answer: (D). ΔT = q / (mc). With q and m the same for every sample, the smallest c gives the largest ΔT. Copper has the smallest c (0.385), so it warms by 500 ÷ (50.0 × 0.385) = 26.0 °C, to 46.0 °C.
- (A) Water has the largest c, so it warms least: only 2.39 °C.
- (B) Aluminium warms by 11.1 °C; its c is more than twice copper’s.
- (C) Iron warms by 22.3 °C, close to copper but less, because its c (0.449) is larger.
Question 3 (multiple choice · core)
Silver has a specific heat capacity of 0.233 J g⁻¹ °C⁻¹. What is its molar heat capacity?
- (A) 2.16 × 10⁻³ J mol⁻¹ °C⁻¹
- (B) 0.233 J mol⁻¹ °C⁻¹
- (C) 25.1 J mol⁻¹ °C⁻¹
- (D) 463 J mol⁻¹ °C⁻¹
Answer and explanation
Answer: (C). One mole of silver has a mass of 107.87 g, so C_m = c × M = 0.233 J g⁻¹ °C⁻¹ × 107.87 g mol⁻¹ = 25.1 J mol⁻¹ °C⁻¹. The grams cancel, which confirms the setup.
- (A) divides c by M; the units would be J g⁻² mol °C⁻¹, which is meaningless.
- (B) copies the specific value and only changes the unit.
- (D) divides M by c (107.87 ÷ 0.233).
Question 4 (multiple choice · core)
A student dissolves a solid in water in a coffee-cup calorimeter. The temperature of the mixture rises from 21.0 °C to 27.5 °C. Which statement is correct?
- (A) Dissolving is endothermic, because the solution absorbed heat.
- (B) Dissolving is exothermic, and q for the dissolving process is negative.
- (C) Dissolving is exothermic, and q for the dissolving process is positive.
- (D) No conclusion is possible without knowing the mass of the solid.
Answer and explanation
Answer: (B). The solution warmed by 6.5 °C, so it gained energy (q(solution) positive). That energy came from the dissolving process, so q(dissolving) = −q(solution) is negative: dissolving is exothermic.
- (A) describes the solution correctly but draws the wrong conclusion about the process. The thermometer is in the surroundings.
- (C) has the right label but the wrong sign for the process.
- (D) The mass is needed to calculate how much energy per gram or per mole, but the direction of energy flow follows from the temperature change alone.
Question 5 (calculation · core)
A 120.0 g block of iron at 95.0 °C is placed in 200.0 g of water at 20.0 °C in an insulated calorimeter. Assume the calorimeter absorbs no energy.
(a) Write an equation, based on conservation of energy, that links the heat transferred by the iron and by the water. (b) Calculate the final temperature. (c) Explain why the final temperature is much closer to 20.0 °C than to 95.0 °C.
Worked solution and suggested Marlbridge rubric
(a) q(iron) + q(water) = 0, so m(Fe) c(Fe) (T_f − 95.0) + m(w) c(w) (T_f − 20.0) = 0.
(b) Heat capacities: iron 120.0 × 0.449 = 53.88 J °C⁻¹; water 200.0 × 4.18 = 836.0 J °C⁻¹. 53.88 (T_f − 95.0) + 836.0 (T_f − 20.0) = 0 889.88 T_f = 53.88 × 95.0 + 836.0 × 20.0 = 21 838.6 T_f = 24.5 °C
Check: iron loses 53.88 × (95.0 − 24.541) = 3796 J; water gains 836.0 × (24.541 − 20.0) = 3796 J. Energy lost = energy gained.
(c) The water’s heat capacity (836.0 J °C⁻¹) is about 15.5 times the iron’s (53.88 J °C⁻¹): it has more mass and a much larger specific heat capacity. The same 3796 J changes the water’s temperature by only 4.5 °C but the iron’s by 70.5 °C.
| Point | What earns it |
|---|---|
| 1 | Correct energy-conservation equation (or “heat lost by iron = heat gained by water”) |
| 1 | Correct substitution, with ΔT written as final − initial for each part |
| 1 | T_f = 24.5 °C |
| 1 | (c) Links the result to the water’s much larger mc (heat capacity), not just “more water” |
A final temperature of 57.5 °C (the midpoint) earns no credit for (b): it assumes equal heat capacities.
Question 6 (constructed response · core)
To find the heat capacity of a calorimeter, a student puts 50.0 g of water at 20.4 °C into the calorimeter and waits until the calorimeter is at 20.4 °C too. The student then quickly adds 50.0 g of water at 45.2 °C, stirs, and records a final temperature of 32.1 °C.
(a) What final temperature would be expected if the calorimeter absorbed no energy? Explain. (b) Calculate the heat lost by the warm water and the heat gained by the cool water. (c) Calculate the heat capacity of the calorimeter, in J °C⁻¹, to an appropriate number of significant figures. (d) Identify one assumption in your calculation.
Worked solution and suggested Marlbridge rubric
(a) 32.8 °C, halfway between 20.4 and 45.2 °C. With equal masses of the same substance, the warm water’s fall must equal the cool water’s rise.
(b) Warm water: ΔT = 32.1 − 45.2 = −13.1 °C; q = 50.0 × 4.18 × (−13.1) = −2737.9 J (lost 2.74 kJ). Cool water: ΔT = 32.1 − 20.4 = +11.7 °C; q = 50.0 × 4.18 × 11.7 = +2445.3 J (gained 2.45 kJ).
(c) Energy is conserved: q(warm) + q(cool) + q(calorimeter) = 0, so q(calorimeter) = 2737.9 − 2445.3 = 292.6 J. The calorimeter started at 20.4 °C, so its ΔT is 11.7 °C. C(calorimeter) = 292.6 J ÷ 11.7 °C = 25.01 J °C⁻¹ = 25 J °C⁻¹. Two significant figures is appropriate: the 292.6 J comes from subtracting two similar numbers. An error of just 0.1 °C in one reading changes either q by about 21 J, so the difference is uncertain by tens of joules.
(d) Any one: no energy is lost to the air or bench; the water’s specific heat capacity is 4.18 J g⁻¹ °C⁻¹ at these temperatures; the final reading is the true equilibrium temperature of water and calorimeter together; no water evaporates.
| Point | What earns it |
|---|---|
| 1 | (a) 32.8 °C, with the reason (equal masses of water) |
| 1 | (b) Both q values correct, with signs or clear “lost” and “gained” |
| 1 | (c) q(calorimeter) = 292.6 J from energy conservation |
| 1 | (c) 25 J °C⁻¹, using a ΔT of 11.7 °C for the calorimeter, given to 2 or 3 significant figures |
| 1 | (d) A valid assumption |
Accept 25.0 J °C⁻¹ for the fourth point; the comment on precision is good practice but is not required for the mark.
Question 7 (constructed response · stretch)
A student dissolves 3.00 g of a fictional salt GD (M = 120.0 g mol⁻¹) in 40.0 g of water in a coffee-cup calorimeter. The temperature rises from 20.5 °C to 26.3 °C.
(a) State whether dissolving GD is endothermic or exothermic, and justify your answer in terms of the direction of energy flow. (b) Calculate q for the solution and q for the dissolving process. (c) Calculate the energy released per mole of GD, in kJ mol⁻¹. (d) The student predicts: “If I dissolve 6.00 g of GD in the same 40.0 g of water, the temperature will rise by the same 5.8 °C, because it is the same salt.” Evaluate this prediction.
Worked solution and suggested Marlbridge rubric
(a) Exothermic. The temperature of the mixture rose, so the solution gained energy. That energy came from the dissolving process, which therefore released energy.
(b) Mass of solution = 40.0 + 3.00 = 43.00 g; ΔT = 26.3 − 20.5 = 5.8 °C. q(solution) = 43.00 × 4.18 × 5.8 = +1042.5 J (about +1.0 kJ). q(dissolving) = −q(solution) = −1042.5 J.
(c) n(GD) = 3.00 ÷ 120.0 = 0.0250 mol. Energy released per mole = 1042.5 J ÷ 0.0250 mol = 41 700 J mol⁻¹ = 42 kJ mol⁻¹ released (q = −42 kJ per mole of GD).
(d) The prediction is incorrect. Twice the mass is twice the moles, so about twice the energy is released (about 2085 J). That energy goes into a similar mass of solution (46.00 g), so ΔT ≈ 2085 ÷ (46.00 × 4.18) ≈ 10.8 °C, nearly double, not the same. The energy per mole stays the same; the temperature change does not.
| Point | What earns it |
|---|---|
| 1 | (a) Exothermic, because the temperature rise shows energy moved from the dissolving process into the solution |
| 1 | (b) q(solution) ≈ +1.0 × 10³ J using the total mass 43.00 g (accept 40.0 g if the assumption is stated: 970 J) |
| 1 | (b) q(dissolving) is the same size with a negative sign |
| 1 | (c) Moles of GD = 0.0250 mol |
| 1 | (c) 42 kJ mol⁻¹ released (or −42 kJ mol⁻¹), carried forward from (b) |
| 1 | (d) Rejects the prediction: twice the moles release about twice the energy, so ΔT roughly doubles |
Carry forward errors from (b) into (c).
How did you do?
- Q1 or Q2 wrong: re-read “Specific heat capacity and q = mcΔT” and Figure 1 in the study guide.
- Q3 wrong: go back to “Molar heat capacity and the heat capacity of an object” and check the units cancel.
- Q4 or Q7 wrong: remember the thermometer is in the solution: q(process) = −q(solution). See Worked example 3.
- Q5 or Q6 incomplete: practise writing q(lost) + q(gained) = 0 first, as in Worked example 2.
Then tick off the topic checklist.