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Chapter 8: Problem 1119

The internal energy change in a system that has absorbed 2 Kcal of heat and done \(500 \mathrm{~J}\) of work is (A) \(7900 \mathrm{~J}\) (B) \(4400 \mathrm{~J}\) (C) \(6400 \mathrm{~J}\) (D) \(8900 \mathrm{~J}\)

Short Answer

Expert verified
The internal energy change in the system is 7868 J, which is closest to option (A) \(7900 \mathrm{~J}\).

Step by step solution

01

Convert heat absorbed from Kcal to Joules

To convert the heat absorbed from Kcal to Joules, we will use the conversion factor 1 Kcal = 4184 Joules. Heat absorbed = 2 Kcal x 4184 Joules/Kcal = 8368 Joules
02

Apply the first law of thermodynamics

According to the first law of thermodynamics, the change in internal energy of the system ΔU is given by: ΔU = Q - W where Q is the heat absorbed by the system and W is the work done by the system.
03

Calculate the change in internal energy

Now we can plug in the values for Q and W into the formula for ΔU: ΔU = 8368 Joules - 500 Joules ΔU = 7868 Joules
04

Compare the result with the given options

We have found that the change in internal energy ΔU is 7868 Joules. Since it is closest to 7900 Joules, the correct option is: (A) \(7900 \mathrm{~J}\)

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Most popular questions from this chapter

For adiabatic Process which relation is true mentioned below ? \(\gamma=\left\\{\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right\\}\) (A) \(\mathrm{p}^{\gamma} \mathrm{V}=\mathrm{Const}\) (B) \(\mathrm{T}^{\gamma} \mathrm{V}=\mathrm{Const}\) (C) TV \(^{\gamma}=\) Const (D) \(\mathrm{TV}^{\gamma-1}=\) Const

A gas expands from 1 liter to 3 liter at atmospheric pressure. The work done by the gas is about (A) \(200 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(300 \mathrm{~J}\) (D) \(2 \times 10^{5} \mathrm{~J}\)

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A thermodynamic Process in which temperature \(\mathrm{T}\) of the system remains constant throughout Variable \(\mathrm{P}\) and \(\mathrm{V}\) may Change is called (A) Isothermal Process (B) Isochoric Process (C) Isobasic Process (D) None of this
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