Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

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) \(20 \mathrm{~J}\)

Step-by-Step Solution

Answer

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

Step 1: Identify Known Values

The work done by the system, denoted as \( W \), is 10 joules. The heat supplied to the system, denoted as \( Q \), is 40 joules.

Step 2: Understand the Formula

The first law of thermodynamics states that the change in internal energy \( \Delta U \) of a system is the difference between the heat added to the system and the work done by the system. It is given by the formula: \( \Delta U = Q - W \).

Step 3: Substitute Values into the Formula

Using the values from Step 1: \( \Delta U = 40 \text{ J} - 10 \text{ J} \).

Step 4: Calculate the Change in Internal Energy

Perform the subtraction: \( 40 \text{ J} - 10 \text{ J} = 30 \text{ J} \).

Step 5: Verify Units

Ensure that the units of energy are consistent throughout the problem, which they are, as we've used joules for both heat and work.

Key Concepts

Internal Energy Change

In the realm of thermodynamics, internal energy change is a crucial concept that signifies the alteration in a system's total energy. It is symbolized by \( \Delta U \), and it quantifies the difference in energy within a system before and after a process. The internal energy can change due to heat transfer and work done. This concept is fundamental in understanding how energy flows in and out of a system, making it a cornerstone of the first law of thermodynamics.

A positive \( \Delta U \) indicates that the system's internal energy has increased, often due to the addition of energy in the form of heat or reduction in work done by the system. Conversely, a negative \( \Delta U \) implies a decrease in internal energy, possibly because of energy loss through work done or heat release. To compute \( \Delta U \), the first law of thermodynamics provides a simple formula:

• \( \Delta U = Q - W \)

Here, \( Q \) is the heat supplied to the system, and \( W \) is the work done by the system.

For instance, if a system is supplied with 40 joules of heat and does 10 joules of work, the increase in internal energy can be found as follows: \( \Delta U = 40 \text{ J} - 10 \text{ J} = 30 \text{ J} \).

Heat Transfer

Heat transfer is the movement of thermal energy from one body or system to another. This transfer occurs due to a temperature difference, following the principle that heat moves from a hot object to a cooler one until thermal equilibrium is achieved. In thermodynamics, it is crucial to account for how much heat a system gains or loses. Heat flow into a system contributes to increasing its internal energy, whereas heat flow out reduces it.

The amount of heat transferred is represented by \( Q \) in thermodynamic equations and is typically measured in joules, like in the exercise above. Heat transfer can occur through conduction, convection, or radiation, depending on the nature of the bodies involved and their environment.

• Conduction involves direct contact and transfer of heat between molecules.
• Convection involves the movement of fluid, carrying heat with it.
• Radiation involves heat transfer through electromagnetic waves without the need for a medium.

In the exercise, 40 joules of heat were supplied, signifying an energy influx, which contributes to the change in internal energy of the system.

Work Done

Work done, in thermodynamics, relates to the process of energy transfer by any means other than heat. It is the energy required to move something against a force or to compress or expand a system, and is represented by the variable \( W \).

When a system performs work on its surroundings, energy leaves the system. Consequently, work done by a system tends to decrease its internal energy since energy is expended from the system's total capacity. The work done by or on a system can be calculated using the equation:

• For expanding gases: \( W = P \times \Delta V \), where \( P \) is pressure and \( \Delta V \) is volume change.

However, in many introductory physics problems, the work is provided directly or calculated using related data, as in our exercise where 10 joules of work was performed by the system.

Finally, by understanding the work done within the given system, it helps in using the first law of thermodynamics efficiently to determine the other energetic aspects of the system, like its internal energy change. In the exercise, the work done by the system was 10 joules, directly influencing the calculation of the internal energy change as \( \Delta U = 40 \text{ J} - 10 \text{ J} = 30 \text{ J} \).