Thermodynamics quiz worksheet featuring 15 questions on laws of thermodynamics, heat engines, and energy conversion.

Problem Analysis and Solution

The provided image contains a set of multiple-choice questions related to thermodynamics. Below, I will solve each question step by step and explain the reasoning.

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#### Question 1: Which law of thermodynamics is used to understand the concept of energy conservation?

- Options:
- A. Second law
- B. None of these
- C. First law
- D. Zeroth law

Solution:
The First Law of Thermodynamics is the principle of energy conservation. It states that energy cannot be created or destroyed, only transformed from one form to another. Mathematically, it is expressed as:
\[
\Delta U = Q - W
\]
where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.

Thus, the correct answer is:
C. First law

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#### Question 2: In which thermodynamic process is there no flow of heat between the system and the surroundings?

- Options:
- A. Isobaric
- B. Adiabatic
- C. Isothermal
- D. Isochoric

Solution:
An adiabatic process is defined as a thermodynamic process in which there is no transfer of heat between the system and its surroundings. This means \(Q = 0\). During an adiabatic process, any change in the internal energy of the system is due solely to work done on or by the system.

Thus, the correct answer is:
B. Adiabatic

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#### Question 3: If the temperature of the source is increased, the efficiency of a Carnot heat engine

- Options:
- A. First increases then becomes constant
- B. Increases
- C. Decreases
- D. Remains constant

Solution:
The efficiency (\(\eta\)) of a Carnot heat engine is given by:
\[
\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}
\]
where:
- \(T_{\text{source}}\) is the absolute temperature of the source,
- \(T_{\text{sink}}\) is the absolute temperature of the sink.

From this equation, it is clear that increasing the temperature of the source (\(T_{\text{source}}\)) while keeping the sink temperature constant will increase the efficiency of the Carnot engine.

Thus, the correct answer is:
B. Increases

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#### Question 4: An ideal heat engine operates in Carnot's cycle between 227°C and 127°C. It absorbs \(6 \times 10^4 \, \text{J}\) at high temperature. The amount of heat converted into work is

- Options:
- A. \(3.5 \times 10^4 \, \text{J}\)
- B. \(48 \times 10^4 \, \text{J}\)
- C. \(1.6 \times 10^4 \, \text{J}\)
- D. \(1.2 \times 10^4 \, \text{J}\)

Solution:
First, convert the temperatures from Celsius to Kelvin:
\[
T_{\text{source}} = 227^\circ \text{C} + 273 = 500 \, \text{K}
\]
\[
T_{\text{sink}} = 127^\circ \text{C} + 273 = 400 \, \text{K}
\]

The efficiency of the Carnot engine is:
\[
\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} = 1 - \frac{400}{500} = 1 - 0.8 = 0.2
\]

The efficiency is also given by:
\[
\eta = \frac{W}{Q_{\text{source}}}
\]
where \(W\) is the work done and \(Q_{\text{source}}\) is the heat absorbed by the engine. Given \(Q_{\text{source}} = 6 \times 10^4 \, \text{J}\), we can find \(W\):
\[
W = \eta \cdot Q_{\text{source}} = 0.2 \times 6 \times 10^4 = 1.2 \times 10^4 \, \text{J}
\]

Thus, the correct answer is:
D. \(1.2 \times 10^4 \, \text{J}\)

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#### Question 5: What is the source temperature of the Carnot engine in K required to get 70% efficiency? Given sink temperature is 27°C.

- Options:
- A. 90 K
- B. 1000 K
- C. 270 K
- D. 727 K

Solution:
The efficiency of a Carnot engine is given by:
\[
\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}
\]
Given \(\eta = 70\% = 0.7\) and \(T_{\text{sink}} = 27^\circ \text{C} + 273 = 300 \, \text{K}\), we can solve for \(T_{\text{source}}\):
\[
0.7 = 1 - \frac{300}{T_{\text{source}}}
\]
\[
\frac{300}{T_{\text{source}}} = 1 - 0.7 = 0.3
\]
\[
T_{\text{source}} = \frac{300}{0.3} = 1000 \, \text{K}
\]

Thus, the correct answer is:
B. 1000 K

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#### Question 6: 110 Joule of heat is added to a gaseous system, whose internal energy is 40 J. Then the amount of external work done is

- Options:
- A. 150 J
- B. 70 J

Solution:
The First Law of Thermodynamics states:
\[
\Delta U = Q - W
\]
where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.

Given:
- \(Q = 110 \, \text{J}\),
- \(\Delta U = 40 \, \text{J}\).

Rearranging the equation to solve for \(W\):
\[
W = Q - \Delta U = 110 - 40 = 70 \, \text{J}
\]

Thus, the correct answer is:
B. 70 J

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Final Answers:

1. C. First law
2. B. Adiabatic
3. B. Increases
4. D. \(1.2 \times 10^4 \, \text{J}\)
5. B. 1000 K
6. B. 70 J

\boxed{C, B, B, D, B, B}