A technician seals a small amount of water in a steel chamber and warms it. Some water forms steam. Fig. 1 shows the chamber, its temperature probe and the ...

Assessment: Physics 4PH1 | Paper 2 Mock 01 | Written Paper 2 Subject: Physics - 4PH1

Question 1 Report

A technician seals a small amount of water in a steel chamber and warms it. Some water forms steam. Fig. 1 shows the chamber, its temperature probe and the pressure sensor.

liquid watersteamprobesensor© EAGLE BEACON GLOBAL

Table 1 shows readings from the apparatus. The outside air pressure is 101 kPa. The lid has area 0.020 m2.

temperature / degrees Cpressure in chamber / kPa
20101
40107
60113
80119

(a) State the unit used for temperature and the unit used for pressure in Table 1. [2]
(b) Calculate the increase in pressure, in kPa per degrees C, between 20 degrees C and 80 degrees C. [3]
(c) Calculate the force on the lid caused by the pressure difference at 80 degrees C. [3]
(d) Explain why the pressure rises as the temperature of the water and steam rises. [2]
(e) Complete the conversions: 750 cm3 = ........ m3; 1.5 L = ........ cm3. [2]

Answer Details

(a) The table uses degrees Celsius, written °C, for temperature and kPa for pressure. [2 marks]

(b) Find both changes before calculating the rate:

\[\Delta p=119-101=18\text{ kPa}\]
\[\Delta T=80-20=60\,°\text{C}\]
\[\text{pressure increase per }°\text{C}=\frac{18}{60}=0.30\text{ kPa per }°\text{C}\]

The answer is 0.30 kPa per °C. [3 marks]

(c) The pressure outside is 101 kPa, so the pressure difference across the lid at 80 °C is:

\[119-101=18\text{ kPa}=18\,000\text{ Pa}\]

Use \(F=pA\), with pressure in pascals:

\[F=18\,000\times0.020=360\text{ N}\]

The force on the lid is 360 N. [3 marks]

(d) Heating gives the water and steam particles more kinetic energy, so they move faster. They collide with the chamber walls more often and/or with greater force. This increases the force per unit area, so the pressure rises. [2 marks]

(e) Since \(1\text{ cm}^3=1\times10^{-6}\text{ m}^3\):

\[750\text{ cm}^3=750\times10^{-6}=0.000750\text{ m}^3\]

Also, \(1\text{ L}=1000\text{ cm}^3\), so:

\[1.5\text{ L}=1500\text{ cm}^3\]

0.000750 m³; 1500 cm³. [2 marks]

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