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The Particulate Nature of Matter - Thermal Energy Transfers

Grade 12IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The particulate nature of matter states that all matter is composed of tiny particles (atoms or molecules) in constant, random motion.

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Internal Energy (UU) is the sum of the total random kinetic energy (EkE_k) and the total intermolecular potential energy (EpE_p) of the particles in a substance: U=∑Ek+∑EpU = \sum E_k + \sum E_p.

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Temperature (TT) is a measure of the average random kinetic energy of the particles in a substance. In the Kelvin scale, T∝Average EkT \propto \text{Average } E_k.

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Specific Heat Capacity (cc) is the amount of energy required to raise the temperature of a unit mass (1 kg1 \text{ kg}) of a substance by 1 K1 \text{ K} (or 1 ∘C1 \text{ } ^\circ\text{C}).

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Specific Latent Heat (LL) is the amount of energy per unit mass required to change the phase of a substance at a constant temperature. LfL_f refers to fusion (solid to liquid) and LvL_v refers to vaporization (liquid to gas).

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During a phase change, the thermal energy added increases the potential energy of the molecules (breaking intermolecular bonds) while the kinetic energy (and thus temperature) remains constant.

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Thermal energy transfer occurs via three mechanisms: Conduction (particle collisions in solids), Convection (bulk movement of fluids due to density differences), and Radiation (transfer via electromagnetic waves, mainly infrared).

📐Formulae

Q=mcΔTQ = mc\Delta T

Q=mLQ = mL

P=QtP = \frac{Q}{t}

T(K)=t(∘C)+273.15T\text{(K)} = t(^\circ\text{C}) + 273.15

💡Examples

Problem 1:

Calculate the energy required to heat 0.25 kg0.25 \text{ kg} of water from 20∘C20 ^\circ\text{C} to its boiling point at 100∘C100 ^\circ\text{C}, and then completely convert it into steam. (Specific heat capacity of water c=4186 J kg−1K−1c = 4186 \text{ J kg}^{-1}\text{K}^{-1}, Specific latent heat of vaporization Lv=2.26×106 J kg−1L_v = 2.26 \times 10^6 \text{ J kg}^{-1})

Solution:

Qtotal=Qheating+Qphase_changeQ_{total} = Q_{heating} + Q_{phase\_change} Qheating=mcΔT=0.25×4186×(100−20)=83,720 JQ_{heating} = mc\Delta T = 0.25 \times 4186 \times (100 - 20) = 83,720 \text{ J} Qphase_change=mLv=0.25×2.26×106=565,000 JQ_{phase\_change} = mL_v = 0.25 \times 2.26 \times 10^6 = 565,000 \text{ J} Qtotal=83,720+565,000=648,720 J≈6.49×105 JQ_{total} = 83,720 + 565,000 = 648,720 \text{ J} \approx 6.49 \times 10^5 \text{ J}

Explanation:

The total energy is the sum of the energy needed to increase the temperature of the liquid water and the energy needed to break the intermolecular bonds to create steam at a constant temperature.

Problem 2:

An electric heater with a power rating of 1.5 kW1.5 \text{ kW} is used to heat 2.0 kg2.0 \text{ kg} of a liquid. If the temperature of the liquid rises from 15∘C15 ^\circ\text{C} to 40∘C40 ^\circ\text{C} in 30 seconds30 \text{ seconds}, calculate the specific heat capacity of the liquid, assuming no heat loss to the surroundings.

Solution:

E=P×t=1500 W×30 s=45,000 JE = P \times t = 1500 \text{ W} \times 30 \text{ s} = 45,000 \text{ J} Using Q=mcΔTQ = mc\Delta T: 45,000=2.0×c×(40−15)45,000 = 2.0 \times c \times (40 - 15) 45,000=2.0×c×2545,000 = 2.0 \times c \times 25 c=45,00050=900 J kg−1K−1c = \frac{45,000}{50} = 900 \text{ J kg}^{-1}\text{K}^{-1}

Explanation:

Power is the rate of energy transfer. By calculating total energy EE from power and time, we can use the calorimetry equation to solve for the unknown specific heat capacity cc.

Thermal Energy Transfers - Revision Notes & Key Formulas | IB Grade 12 Physics