#heat absorbed

Thermodynamic Terms - System Surroundings and Types of Systems

A system absorbs 800 J of heat at constant volume, and its temperature rises by 10 K. If Cv = 20 J/mol·K, how many moles

At constant volume, Δ U = qv = nCvΔ T. So, 800 = n × 20 × 10, n = 800 / 200 = 4 mol.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems

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Thermodynamic Terms - System Surroundings and Types of Systems

Calculate the heat absorbed when 36 g of ice melts at 0°C. (Δ Hfus = 6.01 kJ/mol, molar mass of H₂O = 18 g/mol)

Moles of ice = 36 / 18 = 2 mol. Heat absorbed = 2 × 6.01 = 12.02 kJ.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems

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Heat Capacity and Calorimetry and Measurement of Enthalpy

Calculate the heat absorbed when 27 g of aluminum is heated from 25°C to 125°C at constant volume. (Cv = 22.4 J/mol·K, m

Moles = 27 / 27 = 1 mol, Δ T = 125 - 25 = 100 K, qv = nCvΔ T = 1 × 22.4 × 100 = 2240 J = 2.24 kJ.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy

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Gibbs Energy and Gibbs Energy Change and Equilibrium

Calculate the heat absorbed when 54 g of water is heated from 20°C to 80°C at constant pressure. (Cp = 75.3 J/mol·K, mol

Moles = 54 / 18 = 3 mol, Δ T = 80 - 20 = 60 K, qp = nCpΔ T = 3 × 75.3 × 60 = 13554 J = 13.55 kJ.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium

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Gibbs Energy and Gibbs Energy Change and Equilibrium

Calculate the heat absorbed when 32 g of sulfur is heated from 30°C to 130°C at constant pressure. (Cp = 22.7 J/mol·K, m

Moles = 32 / 32 = 1 mol, Δ T = 130 - 30 = 100 K, qp = nCpΔ T = 1 × 22.7 × 100 = 2270 J = 2.27 kJ.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium

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First Law of Thermodynamics and Enthalpy and Internal Energy

For a reaction at constant volume, the heat absorbed is equal to:

At constant volume, w = -pΔ V = 0, so from the first law, Δ U = qv, the heat absorbed at constant volume.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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First Law of Thermodynamics and Enthalpy and Internal Energy

A system absorbs 600 J of heat at constant volume, and its internal energy increases by 600 J. What is the work done?

At constant volume, w = -PΔ V = 0. First Law: Δ U = q + w = 600 + 0 = 600 J, so w = 0 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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First Law of Thermodynamics and Enthalpy and Internal Energy

In a process, 300 J of heat is absorbed by a system, and 100 J of work is done on the system. What is Δ U?

Using the first law: Δ U = q + w. Here, q = +300 J (heat absorbed), w = +100 J (work done on system), so Δ U = 300 + 100 = 400 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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Enthalpy Changes - Reaction Formation Combustion and Hess's Law

In a reversible isothermal compression of 2 mol of an ideal gas at 300 K, the work done is 4988 J. What is the heat abso

For isothermal process, Δ U = 0, so q = -w. Compression means w > 0 (4988 J), so q = -4988 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law

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