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Date November 2010 Marks available 3 Reference code 10N.2.HL.TZ0.A2
Level Higher level Paper Paper 2 Time zone Time zone 0
Command term Calculate Question number A2 Adapted from N/A

Question

This question is about fuel for heating.

A room heater burns liquid fuel and the following data are available.

\[\begin{array}{*{20}{l}} {{\text{Density of liquid fuel}}}&{ = 8.0 \times {{10}^2}{\text{ kg}}\,{{\text{m}}^{ - 3}}} \\ {{\text{Energy produced by 1 }}{{\text{m}}^{\text{3}}}{\text{ of liquid fuel}}}&{ = 2.7 \times {{10}^{10}}{\text{ J}}} \\ {{\text{Rate at which fuel is consumed}}}&{ = 0.13{\text{ g}}\,{{\text{s}}^{ - 1}}} \\ {{\text{Temperature at which air enters heater}}}&{ = 12{\text{ °C}}} \\ {{\text{Temperature at which air leaves heater}}}&{ = 32{\text{ °C}}} \\ {{\text{Specific heat capacity of air}}}&{ = 990{\text{ J}}\,{\text{k}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}} \end{array}\]

All the energy output of the room heater raises the temperature of the air moving through it. Use the data to calculate the mass of air that moves through the room heater in one second.

Markscheme

\({\text{temperature change}} = {\text{20 (K)}}\);

\({\text{mass of air}} = \frac{{4400}}{{20 \times 990}}\);

\( = 0.22{\text{ kg}}\);

Award [3] for bald correct answer.

Examiners report

Almost all could recognize that the temperature change was 20 K and could use their earlier value of power output. However the numerical manipulations were sometimes weak (often leading to ludicrously large air masses moving through the heater in 1 second.

Syllabus sections

Core » Topic 8: Energy production » 8.1 – Energy sources
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