Mechanical Engineering Exam > Mechanical Engineering Questions > An insulated tank contains 120L of water at 2...
Start Learning for Free
An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-
- a)3.344 kJ/K
- b)0.204 kJ/K
- c)3.104 kJ/K
- d)6.448 kJ/K
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, ...
The question seems to be incomplete. Please provide the remaining information to answer the question.
Free Test
FREE
| Start Free Test |
Community Answer
An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, ...
To find the total entropy change for this process, we will first calculate the entropy change of the water and the copper block separately and then add them together. Let's break down the solution into steps.
Step 1: Calculate the final temperature of the system
Let Tf be the final temperature of the system. We can find it by using the heat transfer equation:
Q = mcΔT
Where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the heat transfer from the copper block to the water is equal:
m_water * c_water * (Tf - T_water) = m_copper * c_copper * (T_copper - Tf)
Plugging in the given values:
120 L * 997 kg/m³ * 4.18 kJ/kg°C * (Tf - 25°C) = 50 kg * 0.386 kJ/kg°C * (80°C - Tf)
Solving for Tf, we get:
Tf ≈ 27.5°C
Step 2: Calculate the entropy change for water
The entropy change for water can be found using the equation:
ΔS_water = m_water * c_water * ln(Tf/T_water)
Plugging in the given values and the final temperature:
ΔS_water = (120 L * 997 kg/m³) * 4.18 kJ/kg°C * ln(27.5°C / 25°C)
ΔS_water ≈ 0.198 kJ/K
Step 3: Calculate the entropy change for the copper block
The entropy change for the copper block can be found using the equation:
ΔS_copper = m_copper * c_copper * ln(Tf/T_copper)
Plugging in the given values and the final temperature:
ΔS_copper = 50 kg * 0.386 kJ/kg°C * ln(27.5°C / 80°C)
ΔS_copper ≈ 0.006 kJ/K
Step 4: Calculate the total entropy change
Now, we can find the total entropy change by adding the entropy change for water and the entropy change for the copper block:
ΔS_total = ΔS_water + ΔS_copper
ΔS_total = 0.198 kJ/K + 0.006 kJ/K
ΔS_total ≈ 0.204 kJ/K
So, the total entropy change for this process is approximately 0.204 kJ/K.
Step 1: Calculate the final temperature of the system
Let Tf be the final temperature of the system. We can find it by using the heat transfer equation:
Q = mcΔT
Where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the heat transfer from the copper block to the water is equal:
m_water * c_water * (Tf - T_water) = m_copper * c_copper * (T_copper - Tf)
Plugging in the given values:
120 L * 997 kg/m³ * 4.18 kJ/kg°C * (Tf - 25°C) = 50 kg * 0.386 kJ/kg°C * (80°C - Tf)
Solving for Tf, we get:
Tf ≈ 27.5°C
Step 2: Calculate the entropy change for water
The entropy change for water can be found using the equation:
ΔS_water = m_water * c_water * ln(Tf/T_water)
Plugging in the given values and the final temperature:
ΔS_water = (120 L * 997 kg/m³) * 4.18 kJ/kg°C * ln(27.5°C / 25°C)
ΔS_water ≈ 0.198 kJ/K
Step 3: Calculate the entropy change for the copper block
The entropy change for the copper block can be found using the equation:
ΔS_copper = m_copper * c_copper * ln(Tf/T_copper)
Plugging in the given values and the final temperature:
ΔS_copper = 50 kg * 0.386 kJ/kg°C * ln(27.5°C / 80°C)
ΔS_copper ≈ 0.006 kJ/K
Step 4: Calculate the total entropy change
Now, we can find the total entropy change by adding the entropy change for water and the entropy change for the copper block:
ΔS_total = ΔS_water + ΔS_copper
ΔS_total = 0.198 kJ/K + 0.006 kJ/K
ΔS_total ≈ 0.204 kJ/K
So, the total entropy change for this process is approximately 0.204 kJ/K.
|
Explore Courses for Mechanical Engineering exam
|
|
Top Courses for Mechanical EngineeringView all
Question Description
An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer?.
An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer?.
Solutions for An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An insulated tank contains 120L of water at 250C ( ρ = 997 kg/m3, CP = 4.18 kJ/kg0C). A 50kg copper block (CP = 0.386 kJ/kg0C) initially at 800C is dropped into water. Total entropy change for this process will be:-a)3.344 kJ/Kb)0.204 kJ/Kc)3.104 kJ/Kd)6.448 kJ/KCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup