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Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - JEE MCQ


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10 Questions MCQ Test Daily Test for JEE Preparation - Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep)

Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) questions and answers have been prepared according to the JEE exam syllabus.The Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) below.
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Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 1
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A metal ball of mass 0.5 kg falls freely from a height of 10 m and bounces to a height of 5.5 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is?
(Specific heat of metal = 450 J/kg°C )

Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 1

The rise in temperature can be calculated using the formula:

ΔT = (Dissipated Energy) / (Mass * Specific Heat)

Given:

Mass (m) = 0.5 kg
Specific heat (C) = 450 J/kg°C
Dissipated Energy = Initial potential energy - Final potential energy
Initial potential energy = m * g * hinitial
Final potential energy = m * g * hfinal
Where:

g is the acceleration due to gravity (approximately 9.81 m/s²)
hinitial is the initial height (10 m)
hfinal is the final height (5.5 m)
Calculations:
Initial potential energy = 0.5 kg * 9.81 m/s² * 10 m = 49.05 J
Final potential energy = 0.5 kg * 9.81 m/s² * 5.5 m = 26.92875 J
Dissipated Energy = Initial potential energy - Final potential energy = 49.05 J - 26.92875 J = 22.12125 J

Now, calculate the rise in temperature:

ΔT = (22.12125 J) / (0.5 kg * 450 J/kg°C) ≈ 0.098°C

So, the rise in temperature is approximately 0.1°C. Therefore, the correct answer is Option B.

Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 2
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In a microcanonical ensemble, a system A of fixed volume is in contact with a large reservoir B. Then:

Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 2
In a microcanonical ensemble, the total energy (E) of the system A (which includes both A and B) is fixed and conserved. System A cannot exchange energy or particles with the surroundings, so the correct option is:

Option A: A can exchange neither energy nor particles with B.

Therefore, the correct answer is Option A.
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Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 3
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Which of the following is not a thermodynamic state variable?

Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 3
Mass (Option B) is not a thermodynamic state variable. Thermodynamic state variables are properties of a system that depend only on its current state and are independent of its history or how it reached that state. Pressure, temperature, and volume are examples of thermodynamic state variables.
Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 4
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Which of the following statements about the specific heat of a body is/are correct?

  1. It depends upon mass and shape of the body
  2. It is independent of mass and shape of the body
  3. It depends only upon the temperature of the body
Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 4

The correct statement is:

It is independent of mass and shape of the body.
The specific heat of a substance is a property of the material itself and is independent of the mass and shape of the body made from that material. Therefore, Option D is the correct answer.

Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 5
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The thermodynamic state of a gas is expressed using its:
Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 5
The thermodynamic state of a gas is expressed using its pressure, volume, and temperature. This combination of properties is known as the thermodynamic state variables or the state parameters of the gas.

So, the correct answer is Option D.
Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 6
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In gases of diatomic molecules, find the ratio of the two specific heat of gases Cp/Cv
Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 6

The ratio of the two specific heat capacities (Cp/Cv) for diatomic ideal gases is approximately 1.4. This value is a characteristic of diatomic gases and is often denoted by the symbol ? (gamma).

So, the correct answer is Option C.

Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 7
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Which of the following is an intensive variable?
Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 7
Pressure (Option C) is an intensive variable. Intensive variables are properties that are independent of the amount of substance present in a system. Pressure remains the same regardless of the size or mass of the system.

So, the correct answer is Option C.
Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 8
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4 kg of solid material is heated from 15°C to 115°C with the addition of 750 kJ of heat in a furnace. What will be its specific heat?
Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 8

The specific heat (C) of a material can be calculated using the formula:

C = (Q) / (m * ΔT)

Where:

Q is the heat added (750 kJ)
m is the mass of the material (4 kg)
?T is the change in temperature (115°C - 15°C = 100°C)
Calculations:
C = (750 kJ) / (4 kg * 100°C) = 1.875 kJ/kg °C

So, the specific heat of the material is 1.875 kJ/kg °C. Therefore, the correct answer is Option A.

Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 9
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The sprinkling of water reduces slightly the temperature of a closed room because:
Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 9
The sprinkling of water reduces the temperature of a closed room because water has a large latent heat of vaporization. When water evaporates, it absorbs heat from the surroundings, including the room, causing a cooling effect. The energy required for water to change from a liquid to vapor (latent heat of vaporization) is significant, and this heat is taken from the environment, leading to a decrease in temperature.

So, the correct answer is Option B.
Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 10
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Which of the following is an intensive thermodynamic property?
Detailed Solution for Test: Specific heat capacity, Thermodynamic state variables and processes(5 Sep) - Question 10
Pressure (Option A) is an intensive thermodynamic property. Intensive properties are those that do not depend on the size or amount of the substance, and pressure is one such property. It has the same value throughout a system regardless of its size.

So, the correct answer is Option A.
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