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Thermodynamics - Specific Heat Capacity

Grade 11CBSEPhysics

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

šŸ”‘Concepts

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Specific Heat Capacity (ss) is the amount of heat energy required to raise the temperature of a unit mass of a substance by 1Ā K1\text{ K} or 1āˆ˜C1^\circ\text{C}. It is an intensive property and depends on the nature of the material.

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Molar Specific Heat Capacity (CC) is the heat required to raise the temperature of 11 mole of a substance by 1Ā K1\text{ K}. It is divided into two types for gases: Molar specific heat at constant volume (CvC_v) and constant pressure (CpC_p).

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Mayer's Relation: For an ideal gas, the difference between molar specific heat at constant pressure and constant volume is equal to the Universal Gas Constant (RR), expressed as Cpāˆ’Cv=RC_p - C_v = R.

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Degrees of Freedom (ff): The specific heat of a gas depends on its molecular structure. According to the law of equipartition of energy, Cv=f2RC_v = \frac{f}{2}R.

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Principle of Calorimetry: In an isolated system, the heat lost by a hot body is equal to the heat gained by a cold body until thermal equilibrium is reached: Qlost=QgainQ_{\text{lost}} = Q_{\text{gain}}.

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Water has a very high specific heat capacity (approximately 4186Ā JĀ kgāˆ’1Kāˆ’14186\text{ J kg}^{-1}\text{K}^{-1}), which is why it is used as a coolant in automobile radiators and why coastal areas have moderate climates.

šŸ“Formulae

s=Ī”QmĪ”Ts = \frac{\Delta Q}{m \Delta T}

C=1nĪ”QĪ”TC = \frac{1}{n} \frac{\Delta Q}{\Delta T}

Cpāˆ’Cv=RC_p - C_v = R

Cv=f2RC_v = \frac{f}{2}R

Cp=(f2+1)RC_p = \left( \frac{f}{2} + 1 \right)R

Ī³=CpCv=1+2f\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}

šŸ’”Examples

Problem 1:

Calculate the amount of heat required to raise the temperature of 2Ā kg2\text{ kg} of water from 25āˆ˜C25^\circ\text{C} to 100āˆ˜C100^\circ\text{C}. (Specific heat of water s=4200Ā JĀ kgāˆ’1Kāˆ’1s = 4200\text{ J kg}^{-1}\text{K}^{-1})

Solution:

Given: m=2Ā kgm = 2\text{ kg}, s=4200Ā JĀ kgāˆ’1Kāˆ’1s = 4200\text{ J kg}^{-1}\text{K}^{-1}. First, calculate the temperature difference Ī”T=100āˆ’25\Delta T = 100 - 25: 100āˆ’2575\begin{array}{r} 100 \\ -25 \\ \hline 75 \end{array} Using the formula: Ī”Q=msĪ”T\Delta Q = m s \Delta T Ī”Q=2Ɨ4200Ɨ75\Delta Q = 2 \times 4200 \times 75 Ī”Q=8400Ɨ75\Delta Q = 8400 \times 75 Ī”Q=630000Ā J\Delta Q = 630000\text{ J}

Explanation:

The total heat required is calculated by multiplying the mass, the specific heat capacity, and the change in temperature. The result 6.3Ɨ105Ā J6.3 \times 10^5\text{ J} represents the thermal energy absorbed.

Problem 2:

Find the ratio of specific heats (Ī³\gamma) for a monatomic gas, which has 33 degrees of freedom.

Solution:

For a monatomic gas, f=3f = 3. Step 1: Calculate CvC_v: Cv=32RC_v = \frac{3}{2}R Step 2: Calculate CpC_p using Mayer's relation Cp=Cv+RC_p = C_v + R: Cp=32R+R=52RC_p = \frac{3}{2}R + R = \frac{5}{2}R Step 3: Find Ī³\gamma: Ī³=CpCv=5/2R3/2R=53ā‰ˆ1.67\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3} \approx 1.67

Explanation:

The ratio Ī³\gamma determines the adiabatic behavior of the gas. For monatomic gases like Helium or Argon, this value is always approximately 1.671.67.