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Energy - Latent Heat

Grade 9IB

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

🔑Concepts

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Latent heat is the energy absorbed or released by a substance during a change in its physical state (phase) that occurs without a change in temperature.

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Specific Latent Heat (LL) is defined as the amount of thermal energy required to change the state of 1 kg1\text{ kg} of a substance without changing its temperature. Its unit is J/kgJ/kg or J kg−1J\text{ kg}^{-1}.

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Specific Latent Heat of Fusion (LfL_f) refers to the phase change between solid and liquid (melting or freezing). For water, Lf≈3.34×105 J/kgL_f \approx 3.34 \times 10^{5}\ J/kg.

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Specific Latent Heat of Vaporization (LvL_v) refers to the phase change between liquid and gas (boiling or condensing). For water, Lv≈2.26×106 J/kgL_v \approx 2.26 \times 10^{6}\ J/kg.

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On a heating or cooling curve, phase changes are represented by horizontal lines (plateaus). During these periods, the thermal energy is used to overcome intermolecular forces (increasing potential energy) rather than increasing the speed of particles (kinetic energy), which is why the temperature remains constant.

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The total energy QQ required for a phase change depends on the mass mm of the substance and its specific latent heat LL.

📐Formulae

Q=mLQ = mL

L=QmL = \frac{Q}{m}

Total Energy=Mass×Specific Latent Heat\text{Total Energy} = \text{Mass} \times \text{Specific Latent Heat}

💡Examples

Problem 1:

Calculate the thermal energy required to melt 0.5 kg0.5\text{ kg} of ice at 0 ∘C0\ ^\circ C. (Specific latent heat of fusion of ice Lf=334,000 J/kgL_f = 334,000\ J/kg).

Solution:

Q=m×LfQ = m \times L_f Q=0.5 kg×334,000 J/kgQ = 0.5\text{ kg} \times 334,000\ J/kg Q=167,000 JQ = 167,000\ J

Explanation:

To find the energy required for melting, we multiply the mass of the ice by the specific latent heat of fusion. Since the ice is already at 0 ∘C0\ ^\circ C, no energy is spent on changing the temperature.

Problem 2:

A heater supplies 452,000 J452,000\ J of energy to boil away a certain mass of water at 100 ∘C100\ ^\circ C. If the specific latent heat of vaporization of water is 2,260,000 J/kg2,260,000\ J/kg, find the mass of the water that turned into steam.

Solution:

m=QLvm = \frac{Q}{L_v} m=452,000 J2,260,000 J/kgm = \frac{452,000\ J}{2,260,000\ J/kg} m=0.2 kgm = 0.2\text{ kg}

Explanation:

Using the rearranged formula m=QLm = \frac{Q}{L}, we divide the total energy supplied by the specific latent heat of vaporization to determine the mass of the substance that underwent the phase change.

Problem 3:

How much energy is released when 2 kg2\text{ kg} of steam at 100 ∘C100\ ^\circ C condenses into water at 100 ∘C100\ ^\circ C? (Lv=2.26×106 J/kgL_v = 2.26 \times 10^6\ J/kg)

Solution:

Q=m×LvQ = m \times L_v Q=2 kg×2.26×106 J/kgQ = 2\text{ kg} \times 2.26 \times 10^6\ J/kg Q=4.52×106 JQ = 4.52 \times 10^6\ J

Explanation:

Condensation is the reverse of vaporization. The same amount of energy per unit mass is released during condensation as is absorbed during boiling. We use the specific latent heat of vaporization for this calculation.