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An engine receives 15152 J/ sec of heat and produces 5kw of power. The efficiency of the engine is
- a)25%
- b)27%
- c)30%
- d)33%
Correct answer is option 'D'. Can you explain this answer?
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To find the efficiency of the engine, we need to use the formula:
Efficiency = (Useful output energy / Input energy) x 100
Given:
Input energy (heat) = 15152 J/s
Useful output energy (power) = 5 kW = 5000 J/s
Now, let's calculate the efficiency of the engine.
Step 1: Convert the power from kilowatts to joules per second:
1 kW = 1000 J/s
So, 5 kW = 5000 J/s
Step 2: Substitute the values in the formula:
Efficiency = (5000 J/s / 15152 J/s) x 100
Step 3: Simplify the expression:
Efficiency = (0.3295) x 100
Efficiency = 32.95%
Therefore, the efficiency of the engine is approximately 33%, which corresponds to option 'D'.
Explanation:
- The efficiency of an engine represents the ratio of useful output energy (power) to the input energy (heat).
- In this case, the engine receives 15152 J/s of heat and produces 5 kW of power.
- To compare the two values, we need to convert the power from kilowatts to joules per second.
- Substituting the values in the efficiency formula, we can calculate the efficiency as a percentage.
- The calculated efficiency is approximately 32.95%, which can be rounded to 33%.
- Therefore, the correct answer is option 'D'.
Efficiency = (Useful output energy / Input energy) x 100
Given:
Input energy (heat) = 15152 J/s
Useful output energy (power) = 5 kW = 5000 J/s
Now, let's calculate the efficiency of the engine.
Step 1: Convert the power from kilowatts to joules per second:
1 kW = 1000 J/s
So, 5 kW = 5000 J/s
Step 2: Substitute the values in the formula:
Efficiency = (5000 J/s / 15152 J/s) x 100
Step 3: Simplify the expression:
Efficiency = (0.3295) x 100
Efficiency = 32.95%
Therefore, the efficiency of the engine is approximately 33%, which corresponds to option 'D'.
Explanation:
- The efficiency of an engine represents the ratio of useful output energy (power) to the input energy (heat).
- In this case, the engine receives 15152 J/s of heat and produces 5 kW of power.
- To compare the two values, we need to convert the power from kilowatts to joules per second.
- Substituting the values in the efficiency formula, we can calculate the efficiency as a percentage.
- The calculated efficiency is approximately 32.95%, which can be rounded to 33%.
- Therefore, the correct answer is option 'D'.
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An engine receives 15152 J/ sec of heat and produces 5kw of power. The efficiency of the engine isa)25%b)27%c)30%d)33%Correct answer is option 'D'. Can you explain this answer? for SSC 2025 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about An engine receives 15152 J/ sec of heat and produces 5kw of power. The efficiency of the engine isa)25%b)27%c)30%d)33%Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An engine receives 15152 J/ sec of heat and produces 5kw of power. The efficiency of the engine isa)25%b)27%c)30%d)33%Correct answer is option 'D'. Can you explain this answer?.
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