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Date May 2014 Marks available 3 Reference code 14M.2.SL.TZ2.2
Level Standard level Paper Paper 2 Time zone Time zone 2
Command term Outline Question number 2 Adapted from N/A

Question

This question is about energy.

At its melting temperature, molten zinc is poured into an iron mould. The molten zinc becomes a solid without changing temperature.

Outline why a given mass of molten zinc has a greater internal energy than the same mass of solid zinc at the same temperature.

[3]
a.

Molten zinc cools in an iron mould. 

The temperature of the iron mould was 20° C before the molten zinc, at its melting temperature, was poured into it. The final temperature of the iron mould and the solidified zinc is 89° C.

The following data are available.

     Mass of iron mould     \( = 12{\text{ kg}}\)

     Mass of zinc     \( = 1.5{\text{ kg}}\)

     Specific heat capacity of iron     \( = 440{\text{ J}}\,{\text{k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}\)

     Specific latent heat of fusion of zinc     \( = 113{\text{ kJ}}\,{\text{k}}{{\text{g}}^{ - 1}}\)

     Melting temperature of zinc     \( = 420{\text{ }}^\circ {\text{C}}\)

     Using the data, determine the specific heat capacity of zinc.

[4]
b.

Markscheme

same temperature so (average) kinetic energy (of atoms/molecules) the same;

(interatomic) potential energy of atoms is greater for liquid / energy is needed to separate the atoms; } (do not allow “forces are weaker” arguments)

internal energy = potential energy + kinetic energy; (allow BOD for clear algebra)

(so internal energy is greater)

a.

energy lost by freezing zinc \( = 1.5 \times 113000{\text{ }}( = 170000{\text{ J}})\); } (watch for power of ten error)

energy gained by iron \( = 12 \times 440 \times [89 - 20]{\text{ }}( = 364000{\text{ J}})\);

energy lost by cooling solid zinc \( = {\text{195000 (J)}}\);

specific heat capacity of zinc \(\frac{{195000}}{{1.5 \times [420 - 89]}} = 390{\text{ (J}}\,{\text{k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}})\);

Award [3 max] for an answer of 733 (kJ\(\,\)kg–1K–1) (1.5 \( \times \) 113 was used).

or

thermal energy lost by zinc = thermal energy gained by iron;

indication that thermal energy lost by zinc has a latent heat contribution and a specific heat contribution expressed algebraically or numerically;

substitution correct;

answer;

b.

Examiners report

Many candidates scored at least two marks on this straightforward recall of material from the syllabus. In weak answers it was not always clear that the statements of energy referred to molecules, atoms or particles in the solid or liquid. There was also confusion as to whether the melting process involved an increase or decrease in potential energy.

a.

The calculation was well done by many with substantial numbers of correct answers. A common error was a failure to read the units of specific latent heat of the zinc correctly and thus to incur a power of ten error. This lost some though not all of the marks. Solutions that incurred zero or a low score were characterised by ill-presented and unexplained numbers showing that the candidate had little idea of the correct approach to the problem. Examiners expected the answer to this question to begin with “Energy lost by zinc = energy gained by iron” and to proceed step by step from there.

b.

Syllabus sections

Core » Topic 3: Thermal physics » 3.1 – Thermal concepts
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