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A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.
- a)141
- b)163
- c)189
- d)210
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A metal ball of diameter 60mm is initially at 220 °C. The ball is...
D = 60mm = 0.06m
Ti = 220o C, h = 200W/m2K
ρ = 9000Kg/m3
T∝ = 20oC , Cp = 400J/Kgk
K= 400W/mK , t = 90sec
T= ?
Most Upvoted Answer
A metal ball of diameter 60mm is initially at 220 °C. The ball is...
Given:
- Diameter of the metal ball = 60 mm
- Initial temperature of the ball = 220 °C
- Temperature of the air jet = 20 °C
- Heat transfer coefficient = 200 W/m²·K
- Density of the metal ball = 9000 kg/m³
To find:
The temperature of the ball after 90 seconds.
Solution:
Step 1: Calculate the surface area of the ball
The surface area of a sphere can be calculated using the formula:
A = 4πr²
Given the diameter of the ball is 60 mm, the radius can be calculated as:
r = d/2 = 60/2 = 30 mm = 0.03 m
Therefore, the surface area of the ball is:
A = 4π(0.03)² = 0.036π m²
Step 2: Calculate the volume of the ball
The volume of a sphere can be calculated using the formula:
V = (4/3)πr³
Substituting the radius value, we get:
V = (4/3)π(0.03)³ = 0.0009048π m³
Step 3: Calculate the mass of the ball
The mass of the ball can be calculated using the formula:
m = ρV
Given the density of the ball is 9000 kg/m³, substituting the volume value, we get:
m = 9000 * 0.0009048π = 8.1432π kg
Step 4: Calculate the heat transferred
The heat transferred from the ball to the air can be calculated using the formula:
Q = hAΔT
Where:
- Q is the heat transferred
- h is the heat transfer coefficient
- A is the surface area of the ball
- ΔT is the temperature difference between the ball and the air
Substituting the values, we get:
Q = 200 * 0.036π * (220 - 20) = 800π W
Step 5: Calculate the change in temperature
The change in temperature can be calculated using the formula:
Q = mcΔT
Where:
- Q is the heat transferred
- m is the mass of the ball
- c is the specific heat capacity of the material
- ΔT is the change in temperature
Rearranging the formula, we get:
ΔT = Q / (mc)
The specific heat capacity of the material is not given, so let's assume it is 500 J/kg·K.
Substituting the values, we get:
ΔT = 800π / (8.1432π * 500) = 0.039 K
Step 6: Calculate the final temperature
The final temperature can be calculated by subtracting the change in temperature from the initial temperature:
Final temperature = Initial temperature - ΔT = 220 - 0.039 = 219.961 °C
Conclusion:
The temperature of the ball after 90 seconds will be approximately 219.961 °C, which can be rounded to 220 °C. Therefore, the correct answer is
- Diameter of the metal ball = 60 mm
- Initial temperature of the ball = 220 °C
- Temperature of the air jet = 20 °C
- Heat transfer coefficient = 200 W/m²·K
- Density of the metal ball = 9000 kg/m³
To find:
The temperature of the ball after 90 seconds.
Solution:
Step 1: Calculate the surface area of the ball
The surface area of a sphere can be calculated using the formula:
A = 4πr²
Given the diameter of the ball is 60 mm, the radius can be calculated as:
r = d/2 = 60/2 = 30 mm = 0.03 m
Therefore, the surface area of the ball is:
A = 4π(0.03)² = 0.036π m²
Step 2: Calculate the volume of the ball
The volume of a sphere can be calculated using the formula:
V = (4/3)πr³
Substituting the radius value, we get:
V = (4/3)π(0.03)³ = 0.0009048π m³
Step 3: Calculate the mass of the ball
The mass of the ball can be calculated using the formula:
m = ρV
Given the density of the ball is 9000 kg/m³, substituting the volume value, we get:
m = 9000 * 0.0009048π = 8.1432π kg
Step 4: Calculate the heat transferred
The heat transferred from the ball to the air can be calculated using the formula:
Q = hAΔT
Where:
- Q is the heat transferred
- h is the heat transfer coefficient
- A is the surface area of the ball
- ΔT is the temperature difference between the ball and the air
Substituting the values, we get:
Q = 200 * 0.036π * (220 - 20) = 800π W
Step 5: Calculate the change in temperature
The change in temperature can be calculated using the formula:
Q = mcΔT
Where:
- Q is the heat transferred
- m is the mass of the ball
- c is the specific heat capacity of the material
- ΔT is the change in temperature
Rearranging the formula, we get:
ΔT = Q / (mc)
The specific heat capacity of the material is not given, so let's assume it is 500 J/kg·K.
Substituting the values, we get:
ΔT = 800π / (8.1432π * 500) = 0.039 K
Step 6: Calculate the final temperature
The final temperature can be calculated by subtracting the change in temperature from the initial temperature:
Final temperature = Initial temperature - ΔT = 220 - 0.039 = 219.961 °C
Conclusion:
The temperature of the ball after 90 seconds will be approximately 219.961 °C, which can be rounded to 220 °C. Therefore, the correct answer is
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A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer?
Question Description
A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer?.
A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer?.
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A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A metal ball of diameter 60mm is initially at 220 °C. The ball is suddenly cooled by an air jet of 20°C. The heat transfer coefficient is 200 W/m2.K and 9000kg/m3, respectively. The ball temperature (in °C) after 90 seconds will be approximately.a)141b)163c)189d)210Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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