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Physics - Thermal Physics (Specific Heat and Phase Changes)

Grade 10IB

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

🔑Concepts

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Internal Energy: The total internal energy of a system is the sum of the random kinetic energy and the potential energy of its molecules. Temperature (TT) is a measure of the average random kinetic energy.

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Specific Heat Capacity (cc): The amount of thermal energy (QQ) required to raise the temperature of a unit mass (mm) of a substance by one degree Kelvin (1 K1\text{ K}) or Celsius (1∘C1^\circ\text{C}). Unit: J⋅kg−1⋅K−1J \cdot kg^{-1} \cdot K^{-1}.

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Phase Changes: During a phase change, the temperature of the substance remains constant. The thermal energy supplied is used to change the potential energy (breaking/forming molecular bonds) rather than the kinetic energy.

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Specific Latent Heat (LL): The energy per unit mass required to change the phase of a substance at a constant temperature. LfL_f is for fusion (solid to liquid) and LvL_v is for vaporization (liquid to gas). Unit: J⋅kg−1J \cdot kg^{-1}.

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Thermal Equilibrium: When two objects are in thermal contact, heat flows from the object at a higher temperature to the object at a lower temperature until they reach the same temperature (Qlost=QgainQ_{lost} = Q_{gain}).

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Heating/Cooling Curves: These graphs plot Temperature (TT) vs. Time (tt). The plateaus (horizontal sections) represent phase changes where TT is constant.

📐Formulae

Q=mcΔTQ = mc\Delta T

Q=mLQ = mL

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

Qlost=QgainedQ_{\text{lost}} = Q_{\text{gained}}

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

💡Examples

Problem 1:

Calculate the thermal energy required to heat 0.25 kg0.25\text{ kg} of water from 20∘C20^\circ\text{C} to 80∘C80^\circ\text{C}. The specific heat capacity of water is c=4186 J⋅kg−1⋅K−1c = 4186\text{ J}\cdot kg^{-1}\cdot K^{-1}.

Solution:

Q=mcΔTQ = mc\Delta T Q=0.25 kg×4186 J⋅kg−1⋅K−1×(80−20)∘CQ = 0.25\text{ kg} \times 4186\text{ J}\cdot kg^{-1}\cdot K^{-1} \times (80 - 20)^\circ\text{C} Q=0.25×4186×60Q = 0.25 \times 4186 \times 60 Q=62,790 JQ = 62,790\text{ J}

Explanation:

Since there is no phase change (water remains liquid), we use the specific heat capacity formula. The temperature change ΔT\Delta T is 60∘C60^\circ\text{C}.

Problem 2:

How much energy is needed to completely melt 0.5 kg0.5\text{ kg} of ice at 0∘C0^\circ\text{C}? The specific latent heat of fusion of ice is Lf=3.34×105 J⋅kg−1L_f = 3.34 \times 10^5\text{ J}\cdot kg^{-1}.

Solution:

Q=mLfQ = mL_f Q=0.5 kg×(3.34×105 J⋅kg−1)Q = 0.5\text{ kg} \times (3.34 \times 10^5\text{ J}\cdot kg^{-1}) Q=1.67×105 JQ = 1.67 \times 10^5\text{ J}

Explanation:

The process is a phase change from solid to liquid at a constant temperature (0∘C0^\circ\text{C}), so we use the formula for latent heat.

Problem 3:

An electric heater with a power of P=1.5 kWP = 1.5\text{ kW} is used to heat 2.0 kg2.0\text{ kg} of a liquid. If the temperature increases by 30∘C30^\circ\text{C} in 2 minutes2\text{ minutes}, calculate the specific heat capacity (cc) of the liquid.

Solution:

Q=P×t=1500 W×(2×60 s)=180,000 JQ = P \times t = 1500\text{ W} \times (2 \times 60\text{ s}) = 180,000\text{ J} Q=mcΔT  ⟹  c=QmΔTQ = mc\Delta T \implies c = \frac{Q}{m\Delta T} c=180,0002.0×30c = \frac{180,000}{2.0 \times 30} c=180,00060=3000 J⋅kg−1⋅K−1c = \frac{180,000}{60} = 3000\text{ J}\cdot kg^{-1}\cdot K^{-1}

Explanation:

First, find the total energy supplied using E=P×tE = P \times t (converting minutes to seconds). Then, rearrange the specific heat formula to solve for cc.

Thermal Physics (Specific Heat and Phase Changes) Revision - Grade 10 Science IB