Answer:
t = 4.0 min
Explanation:
given data:
diameter of rod = 2 cm
T_1 = 100 degree celcius
Air stream temperature = 20 degree celcius
heat transfer coefficient = 200 W/m2. K
WE KNOW THAT
copper thermal conductivity = k = 401 W/m °C
copper specific heat Cp = 385 J/kg.°C
density of copper = 8933 kg/m3
charateristic length is given as Lc
Biot number is given as
Bi = 0.0025
As Bi is greater than 0.1 therefore lumped system analysis is applicable
so we have
............1
where b is given as
b = 0.01163 s^{-1}
putting value in equation 1
solving for t we get
t = 4.0 min
C.
hope that helped you!!!
The answer is static friction. This is the friction that involves objects that do not move.
Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation: