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A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is
[2013]
- a)519
- b)931
- c)1195
- d)2144
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A steel ball of diameter 60 mm is initially in thermal equilibrium at ...
Diameter of steel ball,
d = 0.06 m
Initial temperature,
Ti = 1030°C
Ambient temperature
T0 = 30°C
Convective heat transfer coefficient
Final temperature of steel ball, T = 430°c
Time : t = ?
...(i)
From (i)
d = 0.06 m
Initial temperature,
Ti = 1030°C
Ambient temperature
T0 = 30°C
Convective heat transfer coefficient
Final temperature of steel ball, T = 430°c
Time : t = ?
...(i) From (i)
Most Upvoted Answer
A steel ball of diameter 60 mm is initially in thermal equilibrium at ...
To solve this problem, we need to use the equation for thermal equilibrium:
Q = mcΔT
Where:
Q = Heat energy transferred
m = Mass of the object
c = Specific heat capacity of the material
ΔT = Change in temperature
First, we need to calculate the mass of the steel ball. To do this, we'll use the formula for the volume of a sphere:
V = (4/3)πr³
Where:
V = Volume of the sphere
r = Radius of the sphere
Given that the diameter of the steel ball is 60 mm, the radius can be calculated as follows:
r = diameter / 2
r = 60 mm / 2
r = 30 mm
Converting the radius to meters:
r = 30 mm * 0.001 m/mm
r = 0.03 m
Using the formula for the volume of a sphere, we can calculate the volume:
V = (4/3)π(0.03 m)³
V = (4/3)π(0.000027 m³)
V = 0.000113097 m³
Now, we can calculate the mass of the steel ball using the formula:
m = density * volume
The density of steel is approximately 7850 kg/m³. Therefore:
m = 7850 kg/m³ * 0.000113097 m³
m = 0.885 kg
Now that we have the mass, we can calculate the heat energy transferred using the equation for thermal equilibrium:
Q = mcΔT
Given that the initial temperature is 1030 °C and the final temperature is 20 °C, the change in temperature is:
ΔT = final temperature - initial temperature
ΔT = 20 °C - 1030 °C
ΔT = -1010 °C
Converting the change in temperature to Kelvin:
ΔT = -1010 °C + 273.15 K/°C
ΔT = -736.85 K
Assuming the specific heat capacity of steel is approximately 460 J/kg·K, we can calculate the heat energy transferred:
Q = (0.885 kg)(460 J/kg·K)(-736.85 K)
Q ≈ -305,124.04 J
The negative sign indicates that heat energy is being lost by the steel ball.
Q = mcΔT
Where:
Q = Heat energy transferred
m = Mass of the object
c = Specific heat capacity of the material
ΔT = Change in temperature
First, we need to calculate the mass of the steel ball. To do this, we'll use the formula for the volume of a sphere:
V = (4/3)πr³
Where:
V = Volume of the sphere
r = Radius of the sphere
Given that the diameter of the steel ball is 60 mm, the radius can be calculated as follows:
r = diameter / 2
r = 60 mm / 2
r = 30 mm
Converting the radius to meters:
r = 30 mm * 0.001 m/mm
r = 0.03 m
Using the formula for the volume of a sphere, we can calculate the volume:
V = (4/3)π(0.03 m)³
V = (4/3)π(0.000027 m³)
V = 0.000113097 m³
Now, we can calculate the mass of the steel ball using the formula:
m = density * volume
The density of steel is approximately 7850 kg/m³. Therefore:
m = 7850 kg/m³ * 0.000113097 m³
m = 0.885 kg
Now that we have the mass, we can calculate the heat energy transferred using the equation for thermal equilibrium:
Q = mcΔT
Given that the initial temperature is 1030 °C and the final temperature is 20 °C, the change in temperature is:
ΔT = final temperature - initial temperature
ΔT = 20 °C - 1030 °C
ΔT = -1010 °C
Converting the change in temperature to Kelvin:
ΔT = -1010 °C + 273.15 K/°C
ΔT = -736.85 K
Assuming the specific heat capacity of steel is approximately 460 J/kg·K, we can calculate the heat energy transferred:
Q = (0.885 kg)(460 J/kg·K)(-736.85 K)
Q ≈ -305,124.04 J
The negative sign indicates that heat energy is being lost by the steel ball.
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A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. Can you explain this answer?
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A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. 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 steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. 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 steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. Can you explain this answer?.
A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. 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 steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. 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 steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. Can you explain this answer?.
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A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m2K. The thermophysical properties of steel are : density ρ = 7800 kg/m2, conductivity k =40 W/mK and specific heat c - 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is[2013]a)519b)931c)1195d)2144Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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