5054/21

Physics 5054/21October/November 2011

Cambridge O-Level · Theory · worked solutions for every part, with the mark scheme

11
questions
75
marks
105
minutes

Topics Energy, Work and Power · Electric Circuits · Forces · Mass, Weight and Density · Thermal Properties of Matter · [Legacy] Temperature and Thermometry · +14 more

Q1Mass, Weight and DensityThermal Properties of MatterFree sample

Fig. 1.1 shows an ice cube at 0C0^\circ\text{C}.

The sides of the cube are of length 0.040 m0.040\text{ m}. Ice at 0C0^\circ\text{C} has a density of 920 kg / m3920\text{ kg / m}^3.

(a)

Calculate

(i)

the mass of the ice cube,

mass = ______

3M
DifficultyMedium-Easy
Worked solution

Working

V=0.040×0.040×0.040=6.4×105 m3V = 0.040 \times 0.040 \times 0.040 = 6.4 \times 10^{-5} \text{ m}^3 m=ρV=920×6.4×105m = \rho V = 920 \times 6.4 \times 10^{-5} m=0.05888 kgm = 0.05888 \text{ kg}

Answer

0.059 kg

Final answer

0.059 kg

Detailed explanation

Walkthrough

First, find the volume of the ice cube. Since it is a cube with side length 0.040 m0.040\text{ m}, the volume is V=0.0403=6.4×105 m3V = 0.040^3 = 6.4 \times 10^{-5}\text{ m}^3. Next, use the density formula ρ=m/V\rho = m/V, rearranged to m=ρVm = \rho V. Substitute the given density 920 kg/m3920\text{ kg/m}^3 and the calculated volume: m=920×6.4×105=0.05888 kgm = 920 \times 6.4 \times 10^{-5} = 0.05888\text{ kg}. Round to two significant figures to match the precision of the given side length, giving 0.059 kg0.059\text{ kg} (or 59 g59\text{ g}).

Key Takeaways

The relationship between density, mass, and volume is ρ=m/V\rho = m/V. The volume of a cube is the side length cubed. Always ensure units are consistent (volume in m3\text{m}^3 when density is in kg/m3\text{kg/m}^3).

Common Mistakes

Forgetting to cube the side length to find the volume (e.g. using 0.040×30.040 \times 3 or 0.04020.040^2). Using inconsistent units, such as mixing cm\text{cm} and m\text{m} without converting. Rounding too early or to too many significant figures when the final answer is required.

Things to Be Careful About

The side length 0.040 m0.040\text{ m} has two significant figures, so the final mass should be given to two significant figures (0.059 kg0.059\text{ kg}). Keep the unrounded value (0.05888 kg0.05888\text{ kg}) in your calculator for use in subsequent parts to avoid round-off errors.

Techniques used
calculate the volume of a cube from its side lengthapply the density formula to find mass
(ii)

the weight of the ice cube.

weight = ______

1M
DifficultyEasy
Worked solution

Working

W=mg=0.05888×10W = mg = 0.05888 \times 10 W=0.5888 NW = 0.5888 \text{ N}

Answer

0.59 N

Final answer

0.59 N

Detailed explanation

Walkthrough

Weight is the force of gravity acting on a mass. It is calculated using the formula W=mgW = mg, where mm is the mass in kg and gg is the gravitational field strength. At O Level, gg is taken as 10 N/kg10\text{ N/kg} unless stated otherwise. Using the unrounded mass from part (i): W=0.05888×10=0.5888 NW = 0.05888 \times 10 = 0.5888\text{ N}. Rounding to two significant figures gives 0.59 N0.59\text{ N}.

Key Takeaways

Weight is a force measured in newtons (N), calculated as mass times gravitational field strength (W=mgW = mg). Mass is a scalar quantity measured in kg, while weight is a vector pointing vertically downwards.

Common Mistakes

Confusing mass and weight. Using g=9.8 N/kgg = 9.8\text{ N/kg} when the syllabus expects 10 N/kg10\text{ N/kg} (though 9.89.8 is often accepted, 1010 is standard for 5054 unless specified). Forgetting to include the unit N in the final answer.

Things to Be Careful About

Use the unrounded mass (0.05888 kg0.05888\text{ kg}) for this calculation to maintain accuracy. The final answer should be rounded to two significant figures (0.59 N0.59\text{ N}) to match the data given in the question.

Techniques used
apply the weight formula W = mg using g = 10 N/kg
(b)

The specific latent heat of fusion of ice is 3.4×105 J / kg3.4 \times 10^5\text{ J / kg}. Calculate the thermal energy (heat) absorbed by the ice cube as it melts.

thermal energy = ______

2M
DifficultyMedium-Easy
Worked solution

Working

Q=ml=0.05888×3.4×105Q = ml = 0.05888 \times 3.4 \times 10^5 Q=20019.2 JQ = 20019.2 \text{ J}

Answer

2.0×1042.0 \times 10^4 J

Final answer

2.0 x 10^4 J

Detailed explanation

Walkthrough

The thermal energy required to change the state of a substance without changing its temperature is given by Q=mlQ = ml, where mm is the mass and ll is the specific latent heat. Substitute the unrounded mass (0.05888 kg0.05888\text{ kg}) and the given specific latent heat of fusion for ice (3.4×105 J/kg3.4 \times 10^5\text{ J/kg}): Q=0.05888×3.4×105=20019.2 JQ = 0.05888 \times 3.4 \times 10^5 = 20019.2\text{ J}. Expressing this in standard form to two significant figures gives 2.0×104 J2.0 \times 10^4\text{ J}. Note that using the rounded mass (0.059 kg0.059\text{ kg}) gives 20060 J20060\text{ J}, which is also accepted as 2.01×104 J2.01 \times 10^4\text{ J} or 2.0×104 J2.0 \times 10^4\text{ J}.

Key Takeaways

Specific latent heat is the energy needed to change the state of 1 kg1\text{ kg} of a substance. The formula Q=mlQ = ml applies to melting (fusion) or boiling (vaporisation). During a change of state, the temperature remains constant while thermal energy is used to break intermolecular bonds.

Common Mistakes

Using the wrong formula (e.g. Q=mcΔθQ = mc\Delta\theta, which is for temperature changes, not state changes). Forgetting to convert the latent heat from J/kg\text{J/kg} correctly or misplacing the decimal point in scientific notation. Rounding the mass too early and getting a slightly different final answer.

Things to Be Careful About

The question asks for the energy absorbed as the ice melts at 0C0^\circ\text{C}. Since there is no temperature change, only the latent heat formula is used. Ensure the final answer is given in joules (J) and to an appropriate number of significant figures (2 or 3).

Techniques used
apply the specific latent heat formula Q = ml

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