A 5cm diameter copper sphere (of density = 8954 kg/m3, specific heat capacity = 0.3831 kJ/kg K) is initially at a uniform temper
ature of 10oC. The sphere is placed inside an environment having a temperature of 200oC. The surface heat transfer coefficient is h = 10 W/(m2K). Under these conditions, we can assume uniform internal temperature for the sphere. What is the temperature inside the sphere after 10 minutes in this environment?
1 answer:
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Answer:
a. 51.84Kj
b. 2808.99 W/m^2
c. 11.75%
Explanation:
Amount of heat this resistor dissipates during a 24-hour period
= amount of power dissipated * time
= 0.6 * 24 = 14.4 Watt hour
(Note 3.6Watt hour = 1Kj )
=14.4*3.6 = 51.84Kj
Heat flux = amount of power dissipated/ surface area
surface area = area of the two circular end + area of the curve surface
= 2.136 *10^-4
Heat flux = = 2808.99
fraction of heat dissipated from the top and bottom surface
=11.75%
Answer:
The radius of curvature is 979 meter
Explanation:
We have given velocity of the canon ball v = 98 m/sec
Acceleration due to gravity
We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity
Acceleration due to gravity is given by , here v is velocity and r is radius of curvature
So
r = 979 meter
So the radius of curvature is 979 meter