N5 Heat & Properties of Matter Flashcards by James Cluckie

What does the particle model (or kinetic theory) tell us about the particles of a substance?

Normally the particles of a substance are in constant random motion.

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Define temperature.

The temperature of a substance is a measure of the mean kinetic energy of its particles

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What can happen to the temperature of a substance when heat is gained or lost?

The temperature of a substance can increase when heat is gained

or decrease when heat is lost.

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Define specific heat capacity.

Specific heat capacity is the amount of heat energy required to change the temperature of 1 kg of a material by 1 °C.

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E_h = cmΔT

(Define symbols and units)

E_h - Heat energy (J)

c - specific heat capacity (J kg^-1 °C^-1 - see data sheet)

m - mass (kg)

ΔT - Change in temperature (°C)

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Example

Calculate the amount of heat energy required to boil a kettle of water (2 kg) from a room temperature of 20 °C.

E<sub>h</sub> = ?
c = 4180 J kg<sup>-1</sup> °C<sup>-1</sup> (from data sheet)
m = 2 kg
ΔT = 100 - 20 = 80 °C

*E<sub>h</sub> = cmΔT
E<sub>h</sub> = 4180 x 2 x 80
E<sub>h</sub> = 669 000 J (to 3sf)*

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What happens to a substance when heat is gained or lost at its melting or boiling point?

The substance will change state.

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Describe what happens to the particles of a substance when it changes state.

The bonds between particles are either loosened or strengthened (meaning the particles gain or lose potential energy).

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What happens to the temperature during a change of state?

There is no change in temperature during a change of state.

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E_h = ml

(Define symbols and units)

E_h - Heat energy (J)

m - mass (kg)

l - specific latent heat (J kg^-1 - see data sheet)

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Example

How much water will evaporate when 500 kJ of heat is applied?

E_h = 500 kJ = 500 x 10³ J
m - ?
l = 22·6 x 10⁵ J kg^-1 (from data sheet)

E_h = ml
500 x 10³ = m x 22·6 x 10⁵
m x 22·6 x 10⁵ = 500 x 10³
m = 500 x 10³/22·6 x 10⁵
m = 0·221 kg

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The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled A?

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is increasing temperature as a solid.

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The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled B?

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is changing state from solid to liquid.

(or the substance is melting)

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The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled C?

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is increasing tempearture as a liquid.

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The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled D?

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is changing state from liquid to gas.

(or the substance is boiling)

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The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled E?

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is increasing temperature as a gas.

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The heating curve (graph of temperature versus time) for a substance is shown.

What is the melting point of the substance?

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest. Thus it melts during section B.)

The melting point is 60 °C.

The heating curve (graph of temperature versus time) for a substance is shown.

What is the boiling point of the substance?

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest. Thus it boils during section D.)

The boiling point is 130 °C.

What is the definition of the term pressure?

Pressure is defined as the force per unit area.

p = F/A

(Define symbols and units)

p - pressure (Pa)

F - Force (N)

A - Area (m²)

Example

Calculate the pressure exerted by a 20 kg box with a base measuring 20 cm by 30 cm.

p = ?
F = W = mg = 20 x 9.8 = 196 N
A = 20cm x 30xm = 0.2m x 0.3m = 0.06 m<sup>2</sup>

*p = F/A
p = 196/0.06
p = 3270 Pa (to 3 sig fig)*

Explain how a snowshoe works

(in terms of pressure, force and area).

A snowshow spreads a force over a larger area - this causes the pressure on the snow to decrease.

Account for the pressure of a gas, in terms of the particle model or kinetic theory.

The particles of a gas are in constant random motion
and will collide with their surroundings.
Each collision contributes a small amount of force.
The pressure = Total Force/Area.

State the relationship between pressure and volume for a fixed mass of gas at constant temperature.

The pressure of a fixed mass of gas at constant temperature is inversely proportional to its volume.

*p₁V₁ = p₂V₂* (Define symbols and units)

p - pressure (Pa) V - Volume (m³) Other units can be used if consistent on both sides.

Example A gas container has a volume of 20 cm³ at atmospheric pressure (1.01 x 10⁵ Pa). Calculate the pressure inside when it expands to a volume of 1000 cm³.

``` p₁ = 1.01 x 10⁵ Pa V₁ = 20 cm³ p₂ = ? V₂ = 1000 cm³ ``` *p₁V₁ = p₂V₂* *1.01 x 10⁵ x 20 = p₂ x 1000 p₂ x 1000 = 1.01 x 10⁵ x 20 p₂ = 1.01 x 10⁵ x 20 / 1000 p₂ = 2.02 x 10³ Pa*

State the relationship between **pressure** and **temperature** for a fixed mass of gas at constant volume.

The **pressure** of a fixed mass of gas at constant volume is **directly proportional** to its **temperature measured in Kelvin.**

Explain what is meant by the absolute zero of temperature.

Absolute zero is the temperature (0 K) at which an ideal gas exerts zero pressure and occupies zero volume. This is because it is the temperature at which the gas particles have no kinetic energy and stop moving.

Describe how to convert a temperature in °C into K.

(Reminder: 0 K = -273 °C) To convert a temperature from °C into K, **add 273**.

Describe how to convert a temperature in K into °C.

(Reminder: 0 K = -273 °C) To convert a temperature from K into °C, **subtract 273**.

*p₁/T₁ = p₂/T₂* (Define symbols and units)

p - pressure (Pa) T - temperature (**must** be K) Other units for p can be used if consistent on both sides.

An aerosol can at room temperature (20 °C) contains gas at a pressure of 1·5 x 10⁵ Pa. Calculate the temperature when the can is sprayed and the gas emitted is at atmospheric pressure (1·01 x 10⁵ Pa).

``` p₁ = 1·5 x 10⁵ Pa T₁ = 20 °C = 20+273 = 293 K p₂ = 1·01 x 10⁵ Pa T₂ = ? ``` *p₁/T₁ = p₂/T₂ 1·5 x 10⁵/293 = 1·01 x 10⁵/T₂ Cross Multiply 1·5 x 10⁵ x T₂ = 1·01 x 10⁵ x 293 T₂ = 1·01 x 10⁵ x 293 / 1·5 x 10⁵ T₂ = 197 K (NB leave in K unless the question specifically asks for the answer in °C)*

State the relationship between **volume** and **temperature** for a fixed mass of gas at constant pressure.

The **volume** of a fixed mass of gas at constant pressure is **directly proportional** to its **temperature measured in Kelvin****.**

*V₁/T₁ = V₂/T₂* (Define symbols and units)

V - volume (m³) T - temperature (**must** be K) Other units for V can be used if consistent on both sides.

A gas is held in a flexible container so that its pressure is constant. It has an inital volume of 400 m³ and an inital temperature of 40 °C. What temperature would be required to double this volume?

``` V₁ = 400 m³ T₁ = 40 °C = 40 + 273 = 313 K V₂ = 2 x 400 = 800 m³ T₂ = ? ``` *V₁/T₁ = V₂/T₂ 400/313 = 800/T₂ Cross multiply 400 x T₂ = 800 x 313 T₂ = 800 x 313 / 400 T₂ = 626 K* *(NB leave in K unless the question specifically asks for the answer in °C)*

``` *pV/T = constant p₁V₁/T₁ = p₂V₂/T₂* ``` (Define symbols and units)

p - pressure (Pa) V - volume (m³) T - temperature (**must** be K) Other units for p and V can be used if consistent on both sides.

A 250 ml aerosol can at room temperature (20 °C) contains gas at a pressure of 5·0 x 10⁵ Pa. Calculate the temperature when the can is sprayed to empty and 1000 ml of gas is emitted at atmospheric pressure (1·01 x 10⁵ Pa).

``` p₁ = 5·0 x 10⁵ Pa V₁ = 250 ml T₁ = 20 °C = 20 + 273 = 293 K p₂ = 1·01 x 10⁵ Pa V₂ = 1000 ml T₂ = ? ``` *p₁V₁/T₁ = p₂V₂/T₂ 5·0 x 10⁵ x 250/293 = 1·01 x 10⁵ x 1000/T₂ Cross Multiply 5·0 x 10⁵ x 250 x T₂ = 1·01 x 10⁵ x 1000 x 293 T₂ = 1·01 x 10⁵ x 1000 x 293 / 5·0 x 10⁵ x 250 T₂ = 237 K (NB leave in K unless the question specifically asks for the answer in °C)*

Explain, in terms of **particles**, what happens to the **pressure** of a gas when its **volume is increased**? (The mass and temperature of the gas are fixed.)

* When the **volume** of a gas **is** **increased**, the particles have more space to move around in * The particles **collide** with the container walls **less frequently** * The total force on the walls decreases and because pressure = Force/Area, the **pressure decreases**.

Explain, in terms of **particles**, what happens to the **pressure** of a gas when its **temperature is increased**? (The mass and volume of the gas are fixed.)

* When the **temperature** of a gas **is** **increased**, the particles gain kinetic energy i.e. move faster * The particles **collide** with the container walls **more frequently** and with greater individual force * The total force on the walls increases and because pressure = Force/Area, the **pressure increases**.

Explain, in terms of **particles**, what happens to the **volume** of a gas when its **temperature is increased**? (The mass and pressure of the gas are fixed.)

* When the **temperature** of a gas **is** **increased**, the particles gain kinetic energy i.e. move faster * The particles **collide** with the container walls **more frequently** and with greater individual force * The total force on the walls increases but the container is flexible so it expands i.e. the **volume increases**.