area = 1 cm^2 * 10^-4 m^2/cm^2 = 10^-4 m^2
length = 0.20 m
delta T = -120
dT/dx = -120/0.20 = - 600 deg C/m
Q = 380 W m^-1 degC^-1 * 10^-4 m^2 * 600 degC/m
= 380*10^-4*600 Watts
= 228 watts
dT/dx = -600
if x = 8 cm = 0.08 m
T = 120 - 600(.08)
= 72 degC
other way
(8/20)120= 48 deg lost
120 - 48 = 72 deg C luckily
A lagged copper rod has a uniform
cross-sectional area of 1.0cm3 and length 20.0cm.
when steady state is attained, the temperature of one
end of the rod is 120degree Celsius and other end is
0degree Celsius. Calculate; (1)the temperature gradient
(2).rate of heat flow
(3).The temperature at a point 8.0cm from the hot end. (the thermal conductivity of copper =380Wm-1k-1)
3 answers
Thanks bro... I appreciate a lot
You are welcome.
1 answer