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Physics: Thermal Physics - Specific Heat Capacity and Latent Heat

Grade 8IB

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

🔑Concepts

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Internal energy is the total energy stored by the particles that make up a system. It is the sum of the total Kinetic Energy (associated with temperature) and Potential Energy (associated with the bonds/state of the particles).

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Specific Heat Capacity (cc) is defined as the amount of thermal energy required to raise the temperature of 1 kg1\ kg of a substance by 1∘C1^{\circ}C (or 1 K1\ K). Its unit is J/kg∘CJ/kg^{\circ}C or J kg−1K−1J\ kg^{-1}K^{-1}.

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Different substances have different heat capacities; for instance, water has a very high specific heat capacity (4200 J/kg∘C4200\ J/kg^{\circ}C), meaning it heats up and cools down slowly.

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Specific Latent Heat (LL) is the energy required to change the state (phase) of 1 kg1\ kg of a substance without any change in temperature.

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Specific Latent Heat of Fusion (LfL_f) is the energy needed to change a substance from solid to liquid at its melting point.

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Specific Latent Heat of Vaporization (LvL_v) is the energy needed to change a substance from liquid to gas at its boiling point.

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On a heating curve, the 'flat' horizontal sections represent phase changes where the temperature remains constant because the energy is being used to break intermolecular bonds (increasing Potential Energy) rather than increasing the speed of particles (Kinetic Energy).

📐Formulae

Q=mcΔTQ = mc\Delta T

Q=mLQ = mL

ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}

c=QmΔTc = \frac{Q}{m\Delta T}

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

💡Examples

Problem 1:

A copper block of mass 0.8 kg0.8\ kg is heated from 25∘C25^{\circ}C to 75∘C75^{\circ}C. If the specific heat capacity of copper is 390 J/kg∘C390\ J/kg^{\circ}C, calculate the thermal energy absorbed.

Solution:

Mass m=0.8 kgm = 0.8\ kg Specific heat c=390 J/kg∘Cc = 390\ J/kg^{\circ}C Change in temperature ΔT=75−25=50∘C\Delta T = 75 - 25 = 50^{\circ}C

Q=mcΔTQ = mc\Delta T Q=0.8×390×50Q = 0.8 \times 390 \times 50 Q=15,600 JQ = 15,600\ J

Explanation:

The formula Q=mcΔTQ = mc\Delta T is used because there is a change in temperature but no change in state. We multiply the mass, the constant for copper, and the temperature difference.

Problem 2:

How much energy is required to completely vaporize 0.2 kg0.2\ kg of ethanol at its boiling point? (Specific latent heat of vaporization of ethanol Lv=8.46×105 J/kgL_v = 8.46 \times 10^5\ J/kg)

Solution:

Mass m=0.2 kgm = 0.2\ kg Latent heat Lv=846,000 J/kgL_v = 846,000\ J/kg

Q=mLvQ = mL_v Q=0.2×846,000Q = 0.2 \times 846,000 Q=169,200 JQ = 169,200\ J

Explanation:

The formula Q=mLQ = mL is used because the substance is at its boiling point and undergoing a phase change (liquid to gas) at a constant temperature.

Problem 3:

An electric heater provides 5000 J5000\ J of energy to a 0.5 kg0.5\ kg sample of an unknown metal, causing its temperature to rise by 25∘C25^{\circ}C. Find the specific heat capacity of the metal.

Solution:

c=QmΔTc = \frac{Q}{m\Delta T} c=50000.5×25c = \frac{5000}{0.5 \times 25} c=500012.5c = \frac{5000}{12.5} c=400 J/kg∘Cc = 400\ J/kg^{\circ}C

Explanation:

By rearranging the specific heat formula to solve for cc, we divide the total energy by the product of mass and temperature change.