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Energy - Thermal Energy Transfer

Grade 9IB

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

🔑Concepts

Thermal energy always transfers from a region of higher temperature to a region of lower temperature until thermal equilibrium is reached.

Conduction: The process where thermal energy is passed through a substance by the vibration of atoms and the movement of free electrons. It is most effective in solids, especially metals, which have a high density of free electrons (ee^-).

Convection: Thermal energy transfer in fluids (liquids and gases) caused by differences in density. When a fluid is heated, it expands, its density decreases (ho=mV ho = \frac{m}{V}), and it rises, creating a convection current.

Radiation: The transfer of energy by infrared (IR) electromagnetic waves. Unlike conduction and convection, radiation does not require a medium and can travel through a vacuum. Surface properties affect this: dull, black surfaces are the best absorbers and emitters, while shiny, silver surfaces are the best reflectors.

Specific Heat Capacity (cc): The amount of energy required to raise the temperature of 1 kg1\text{ kg} of a substance by 1 C1\text{ }^\circ\text{C} (or 1 K1\text{ K}). It is measured in J kg1 C1\text{J kg}^{-1}\text{ }^\circ\text{C}^{-1}.

Specific Latent Heat (LL): The energy required to change the state of 1 kg1\text{ kg} of a substance without a change in temperature. LfL_f is for fusion (melting/freezing) and LvL_v is for vaporization (boiling/condensing).

📐Formulae

Q=mcΔTQ = mc\Delta T

Q=mLQ = mL

Efficiency=(Useful Energy OutputTotal Energy Input)×100%\text{Efficiency} = \left( \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \right) \times 100\%

ρ=mV\rho = \frac{m}{V}

💡Examples

Problem 1:

A 0.5 kg0.5\text{ kg} block of aluminum is heated from 25 C25\text{ }^\circ\text{C} to 75 C75\text{ }^\circ\text{C}. Calculate the thermal energy absorbed by the block. (Specific heat capacity of aluminum c=900 J kg1 C1c = 900\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1})

Solution:

Q=0.5 kg×900 J kg1 C1×(75 C25 C)Q = 0.5\text{ kg} \times 900\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1} \times (75\text{ }^\circ\text{C} - 25\text{ }^\circ\text{C}) Q=0.5×900×50Q = 0.5 \times 900 \times 50 Q=22,500 JQ = 22,500\text{ J}

Explanation:

We use the specific heat capacity formula Q=mcΔTQ = mc\Delta T. The mass mm is 0.5 kg0.5\text{ kg}, the specific heat cc is 900 J kg1 C1900\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1}, and the change in temperature ΔT\Delta T is 50 C50\text{ }^\circ\text{C}.

Problem 2:

How much energy is required to completely melt 0.2 kg0.2\text{ kg} of ice at 0 C0\text{ }^\circ\text{C}? (Specific latent heat of fusion of ice Lf=3.34×105 J kg1L_f = 3.34 \times 10^5\text{ J kg}^{-1})

Solution:

Q=mLfQ = mL_f Q=0.2 kg×3.34×105 J kg1Q = 0.2\text{ kg} \times 3.34 \times 10^5\text{ J kg}^{-1} Q=66,800 JQ = 66,800\text{ J}

Explanation:

Since the temperature remains constant during a state change (melting), we use the latent heat formula Q=mLQ = mL. LfL_f is used because the process is fusion (melting).

Thermal Energy Transfer - Revision Notes & Key Formulas | IB Grade 9 Science