Heat & Thermodynamics

Temperature and Heat

5th Year · 6th Year (Leaving Cert)

✓By the end of this lesson students will be able to distinguish between temperature and heat.
✓By the end of this lesson students will be able to convert between the Celsius and Kelvin temperature scales.
✓By the end of this lesson students will be able to define and apply the concept of specific heat capacity.
✓By the end of this lesson students will be able to define and apply the concept of specific latent heat.
✓By the end of this lesson students will be able to solve problems involving heat transfer using relevant formulae.

Key concepts

Temperature

Temperature is a measure of the average kinetic energy of the particles within a substance. It determines the direction of heat flow between two bodies in thermal contact. The SI unit for temperature is the Kelvin (K).

Heat

Heat is a form of energy that is transferred from a hotter body to a colder body due to a temperature difference. It is measured in joules (J). Heat is energy in transit, not a property stored within a body.

Celsius vs Kelvin Scales

The Celsius scale (°C) is a commonly used practical temperature scale. The Kelvin scale (K) is the absolute temperature scale, where 0 K (absolute zero) represents the lowest possible temperature, at which particles have minimum kinetic energy. A change of 1°C is numerically equal to a change of 1 K.

T(K) = θ(°C) + 273.15 (For Leaving Cert calculations, 273 is often used for simplicity unless higher precision is specified.)

Specific Heat Capacity (c)

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 kg of that substance by 1 K (or 1°C). Its unit is joules per kilogram per Kelvin (J kg⁻¹ K⁻¹) or joules per kilogram per degree Celsius (J kg⁻¹ °C⁻¹).

Q = mcΔθ

Latent Heat

Latent heat is the heat energy absorbed or released by a substance during a phase change (e.g., melting, boiling, freezing, condensing) without a change in its temperature. This energy is used to change the potential energy of the particles, breaking or forming intermolecular bonds.

Specific Latent Heat of Fusion (l_f)

The specific latent heat of fusion is the amount of heat energy required to change 1 kg of a substance from solid to liquid (or vice versa) at its melting point without a change in temperature. Its unit is joules per kilogram (J kg⁻¹).

Q = ml_f

Specific Latent Heat of Vaporisation (l_v)

The specific latent heat of vaporisation is the amount of heat energy required to change 1 kg of a substance from liquid to gas (or vice versa) at its boiling point without a change in temperature. Its unit is joules per kilogram (J kg⁻¹).

Q = ml_v

Key facts to remember

1Temperature is a measure of the average kinetic energy of particles, while heat is the transfer of thermal energy.
2The Kelvin scale is the absolute temperature scale, where 0 K is absolute zero.
3To convert Celsius to Kelvin, add 273.15 (or 273 for most exam calculations).
4Specific heat capacity (c) is the heat energy required to change the temperature of 1 kg of a substance by 1 K.
5Latent heat is the energy absorbed or released during a phase change without a change in temperature.
6Specific latent heat of fusion (l_f) is for melting/freezing, and specific latent heat of vaporisation (l_v) is for boiling/condensing.
7The formula for heat transfer due to temperature change is Q = mcΔθ.
8The formula for heat transfer during a phase change is Q = ml (where l is the specific latent heat).

Worked examples

Example 1

A 2.5 kg block of copper is heated from 20 °C to 80 °C. Calculate the heat energy absorbed by the copper. (Specific heat capacity of copper = 390 J kg⁻¹ K⁻¹).

IIdentify the given values:

IIMass, m = 2.5 kg

IIIInitial temperature, θ_initial = 20 °C

IVFinal temperature, θ_final = 80 °C

VSpecific heat capacity of copper, c = 390 J kg⁻¹ K⁻¹

VICalculate the change in temperature (Δθ):

VIIΔθ = θ_final - θ_initial = 80 °C - 20 °C = 60 °C

VIII(Note: A change of 60 °C is equivalent to a change of 60 K, so Δθ = 60 K)

9Apply the formula for heat energy (Q = mcΔθ):

10Q = (2.5 kg) × (390 J kg⁻¹ K⁻¹) × (60 K)

11Q = 58500 J

Answer

The heat energy absorbed by the copper is 58500 J.

Example 2

How much heat energy is required to melt 0.5 kg of ice at 0 °C to water at 0 °C? (Specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹).

IIdentify the given values:

IIMass of ice, m = 0.5 kg

IIISpecific latent heat of fusion of ice, l_f = 3.34 × 10⁵ J kg⁻¹

IVApply the formula for latent heat of fusion (Q = ml_f):

VQ = (0.5 kg) × (3.34 × 10⁵ J kg⁻¹)

VIQ = 1.67 × 10⁵ J

Answer

The heat energy required is 1.67 × 10⁵ J.

During a phase change, the temperature remains constant, so specific heat capacity is not used.

Example 3

Calculate the total heat energy required to convert 0.2 kg of ice at -10 °C to water at 20 °C. (Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹, Specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹, Specific heat capacity of water = 4180 J kg⁻¹ K⁻¹).

IThis problem involves three stages of heat transfer:

IIStage 1: Heat required to raise the temperature of ice from -10 °C to 0 °C.

IIIm = 0.2 kg, c_ice = 2100 J kg⁻¹ K⁻¹, Δθ = 0 °C - (-10 °C) = 10 °C = 10 K

IVQ₁ = mc_iceΔθ = (0.2) × (2100) × (10) = 4200 J

VStage 2: Heat required to melt the ice at 0 °C to water at 0 °C.

VIm = 0.2 kg, l_f = 3.34 × 10⁵ J kg⁻¹

VIIQ₂ = ml_f = (0.2) × (3.34 × 10⁵) = 66800 J

VIIIStage 3: Heat required to raise the temperature of water from 0 °C to 20 °C.

9m = 0.2 kg, c_water = 4180 J kg⁻¹ K⁻¹, Δθ = 20 °C - 0 °C = 20 °C = 20 K

10Q₃ = mc_waterΔθ = (0.2) × (4180) × (20) = 16720 J

11Calculate the total heat energy:

12Total Q = Q₁ + Q₂ + Q₃ = 4200 J + 66800 J + 16720 J = 87720 J

Answer

The total heat energy required is 87720 J.

Always break down complex problems into individual stages of heating/cooling and phase changes.

Common mistakes

✗Confusing temperature (a measure of average kinetic energy) with heat (a form of energy transfer).
✗Forgetting to convert mass from grams to kilograms before using formulae.
✗Using the specific heat capacity of water when the substance is ice (or steam) and vice versa.
✗Omitting the latent heat calculation when a phase change (melting, boiling) occurs.
✗Incorrectly calculating Δθ; always use the change in temperature, not just the final temperature.
✗Not recognising that a change in temperature of 1°C is numerically equal to a change of 1 K.

Exam tips

★Read the question carefully to identify all stages of heat transfer (e.g., heating a solid, melting, then heating the liquid).
★List all given values and constants clearly, including their units, before starting any calculations.
★Show all steps of your working, especially for multi-stage problems, to maximise marks even if the final answer is incorrect.
★Pay close attention to units and ensure consistency throughout your calculations (e.g., use kilograms for mass, joules for energy).

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