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If amount of heat given to a system is \(50 \mathrm{~J}\) and work done on the system is \(15 \mathrm{~J}\), then change in internal energy of the system is (A) \(35 \mathrm{~J}\) (B) \(50 \mathrm{~J}\) (C) \(65 \mathrm{~J}\) (D) \(15 \mathrm{~J}\)

Short Answer

Expert verified
The change in internal energy of the system is \(35 \mathrm{~J}\).

Step by step solution

01

Identify the given values

In this problem, we are given the amount of heat given to the system (Q) as 50 J and the work done on the system (W) as 15 J.
02

Recall the First Law of Thermodynamics

The First Law of Thermodynamics states that "the change in the internal energy (∆U) of a system is equal to the heat added to the system (Q) minus the work done by the system (W)". Mathematically, this can be written as: \(∆U = Q - W\)
03

Substitute the given values

Now, plug in the given values of Q and W into the formula: \(∆U = 50 \mathrm{~J} - 15 \mathrm{~J}\)
04

Calculate the change in internal energy

Calculate the difference between the heat given to the system and the work done on the system: \(∆U = 35 \mathrm{~J}\) So the correct answer is (A) \(\bold{35 \mathrm{~J}}\). The change in internal energy of the system is 35 J.

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Key Concepts

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

Internal Energy
When we talk about the internal energy of a system in thermodynamics, we refer to the total energy contained within a system. This encompasses not only the kinetic energy of molecules moving and vibrating but also any potential energy resulting from molecular interactions. It's a concept that might seem abstract at first, but it plays a pivotal role in understanding how different forms of energy transfer can affect a system.

The internal energy is denoted as \(U\) and it can be thought of as the energy currency of a system. Since it's not directly observable, we often deal with changes in internal energy (\(\Delta U\)), which occur through heat exchange or work being done. In the exercise, we were tasked with calculating this change based on heat input and work done. It's important to remember that internal energy is a state function, meaning its change depends only on the initial and final states of the system, not on how the change occurs.
Thermodynamics Problems
Solving thermodynamics problems typically involves a blend of conceptual understanding and mathematical skills. In the classroom, you'll often deal with exercises that require you to apply the first law of thermodynamics, calculate properties like work and heat, and interpret the implications of energy exchanges.

To tackle these problems effectively, it's crucial to have a clear understanding of the basic principles—like the conservation of energy and the characteristics of thermodynamic processes. From there, the key is to systematically identify given values and known quantities, apply the appropriate formulas, and carefully perform the necessary calculations. For instance, in our example problem, by methodically following the steps, we reached the solution without getting lost in the intricacies of thermodynamic cycles or the molecular details of the system.
Heat and Work
The interplay between heat and work is fundamental in the realm of thermodynamics. Heat (\(Q\)) refers to the transfer of energy due to a temperature difference, while work (\(W\)) accounts for energy transfer attributed to a force acting over a distance.

In our everyday understanding, heat is often associated with warmth, and work relates to physical or mechanical effort. However, in thermodynamics, these concepts have a broader signification. Heat can be transferred into or out of a system, just like work can be done on or by a system. They are pathways by which a system's internal energy can change—in line with the first law of thermodynamics.

Practical Implications of Heat and Work

Concerning the exercise, when heat is given to the system, it tends to increase the internal energy. Conversely, when work is done on the system, it could either increase or decrease its internal energy depending on the direction of work relative to the system. Knowing how to correctly quantify these energy transfers allows for accurate predictions about system behavior in various thermodynamic scenarios.

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Most popular questions from this chapter

A Carnot engine, whose efficiency is \(40 \%\), takes heat from a source maintained at a temperature of \(500 \mathrm{~K}\). It is desired to have an engine of efficiency \(60 \%\). Then, the intake temperature for the same exhaust (sink) temperature must be: (A) Efficiency of Carnot engine cannot be made larger than \(50 \%\) (B) \(1200 \mathrm{~K}\) (C) \(750 \mathrm{~K}\) (D) \(600 \mathrm{~K}\)

Which one of the following statements is incorrect? (A) If positive work is done by a system in a thermodynamic process, its volume must increase. (B) If heat is added to a system, its temperature must increase. (C) A body at \(20^{\circ} \mathrm{C}\) radiates in a room, where room temperature is \(30^{\circ} \mathrm{C}\). (D) If pressure vs temperature graph of an ideal gas is a straight line, then work done by the gas is zero.

Equal amount of same gas in two similar cylinders, \(A\) and \(B\), compressed to same final volume from same initial volume one adiabatically and another isothermally, respectively, then (A) final pressure in \(A\) is more than in \(B\). (B) final pressure in \(B\) is greater than in \(A\). (C) final pressure in both equal. (D) for the gas, value of \(\gamma=\frac{C_{p}}{C_{Y}}\) is required.

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume \(V_{1}\) and contains ideal gas at pressure \(P_{1}\) and temperature \(T_{1}\). The other chamber has volume \(V_{2}\) and contains ideal gas at pressure \(P_{2}\) and temperature \(T_{2}\). If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be: (A) \(\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}\) (B) \(\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}\) (C) \(\frac{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}{P_{1} V_{1}+P_{2} V_{2}}\) (D) \(\frac{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}{P_{1} V_{1}+P_{2} V_{2}}\)

P-V diagram of an ideal gas is as shown in Fig. \(12.10 .\) Work done by the gas in the process \(A B C D\) is (A) \(4 P_{0} V_{0}\) (B) \(2 P_{0} V_{0}\) (C) \(3 P_{0} V_{0}\) (D) \(P_{0} V_{0}\)

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