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The work done by a system is 10 joule, when 40 joule heat is supplied to it. What is the increase in internal energy of system? (a) \(30 \mathrm{~J}\) (b) \(50 \mathrm{~J}\) (c) \(40 \mathrm{~J}\) (d) \(20 \mathrm{~J}\)

Short Answer

Expert verified
The increase in internal energy is 30 J (option a).

Step by step solution

01

Understand the First Law of Thermodynamics

The first law of thermodynamics is expressed by the formula: \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat absorbed by the system, and \( W \) is the work done by the system.
02

Identify the Given Values

We are given that the heat supplied to the system, \( Q = 40 \text{ J} \), and the work done by the system, \( W = 10 \text{ J} \). We need to find the change in internal energy, \( \Delta U \).
03

Substitute the Given Values into the Formula

Using the first law of thermodynamics formula, substitute the known values: \( \Delta U = 40 - 10 \).
04

Calculate the Change in Internal Energy

Perform the calculation: \( \Delta U = 40 - 10 = 30 \text{ J} \).
05

Interpret the Result

The calculated change in internal energy is \( 30 \text{ J} \). Since \( 30 \text{ J} \) matches option (a), this is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Internal Energy
Internal energy is a fundamental concept in thermodynamics, representing the total energy contained within a system. This energy is due to the microscopic activities of particles within the system:
  • Molecular motion
  • Vibrations
  • Intermolecular forces
In simpler terms, internal energy is the stored energy inside a system that can change when the system interacts with its surroundings. For example, when a system absorbs heat or does work on its surroundings, its internal energy changes.
The symbol used for change in internal energy is     \( \Delta U \). It's crucial in understanding various energy transformations and the overall behavior of thermal systems.
Heat Transfer
Heat transfer is the process of energy moving from one place to another due to temperature differences. It is a central concept in thermodynamics because it dictates how energy is exchanged:
  • Between systems
  • Or between a system and its environment
The symbol used to denote the heat added or removed from a system is \( Q \). If the system absorbs heat, this energy contributes to its internal energy, and if the system loses heat, the internal energy decreases.
Understanding heat transfer helps us analyze how thermal systems behave when interacting with their environment. It is one of the three ways a system can exchange energy, along with work and radiative transfer.
Work Done
Work done in thermodynamics refers to energy transferred to or from a system by mechanical means. When we say 'work done by the system,' we mean that the system is using its energy to do something externally. For example:
  • Expanding a gas against an external pressure
  • Moving an object
The symbol for work done is \( W \). When a system does work on its surroundings, its internal energy decreases. In the context of our problem, the system did 10 joules of work, affecting the overall change in internal energy.
The concept of work is crucial for calculating energy changes and understanding the energy balance within a system.
Thermodynamics Formula
The first law of thermodynamics, often referred to as the law of energy conservation, lays the foundation for understanding energy flow within a system. Its formula is expressed as:\[ \Delta U = Q - W \]Here, \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.
This equation indicates that the change in a system's internal energy is the heat energy supplied to it minus the work it performs. It ensures that no energy is lost or destroyed, only transformed. In solving the previous exercise, substituting the known values into this formula allows for calculating the change in internal energy, illustrating this primary principle of thermodynamics in action.

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Most popular questions from this chapter

Calculate \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\) for chloride ion from the following data: \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I} / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HCl}(\mathrm{g})\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-92.4 \mathrm{~kJ}\) \(\mathrm{HCl}(\mathrm{g})+\mathrm{nH}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) \(\Delta \mathrm{H}_{\mathrm{Hyd}}=-74.8 \mathrm{~kJ}\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{f}}\left[\mathrm{H}^{+}\right]=0.0 \mathrm{~kJ}\) (a) \(-189 \mathrm{~kJ}\) (b) \(-167 \mathrm{~kJ}\) (c) \(+167 \mathrm{~kJ}\) (d) \(-191 \mathrm{~kJ}\)

What is the value of \(\Delta \mathrm{E}\), when \(64 \mathrm{~g}\) oxygen is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) at constant volume? \(\left(\mathrm{C}_{\mathrm{v}}\right.\) on an average is \(5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ) (a) \(1500 \mathrm{~J}\) (b) \(1800 \mathrm{~J}\) (c) \(2000 \mathrm{~J}\) (d) \(2200 \mathrm{~J}\)

For a reaction to occur spontaneously (a) \((\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S})\) must be negative (b) \((\Delta \mathrm{H}+\mathrm{T} \Delta \mathrm{S})\) must be negative (c) \(\Delta \mathrm{H}\) must be negative (d) \(\Delta \mathrm{S}\) must be negative

For a reaction at \(300 \mathrm{~K}\), enthalpy and entropy changes are \(-11.5 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) and \(-105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively. What is the change in Gibbs free energy? (a) \(25 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(15 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(20 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

An athlete is given \(100 \mathrm{~g}\) of glucose of energy equivalent to \(1560 \mathrm{~kJ}\). He utilizes \(50 \%\) of this gained energy in the event. In order to avoid storage of energy in the body, calculate the mass of water he would need to perspire. Enthalpy of \(\mathrm{H}_{2} \mathrm{O}\) for evaporation is \(44 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (a) \(346 \mathrm{~g}\) (b) \(316 \mathrm{~g}\) (c) \(323 \mathrm{~g}\) (d) \(319 \mathrm{~g}\)

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