How much energy in joule is added to a 12 g of sample of aluminum (c=0.897 J/g ◦C) to raise the temperature from 20 ◦C to 45 ◦C?
1 answer:
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Answer:
heat loss per 1-m length of this insulation is 4368.145 W
Explanation:
given data
inside radius r1 = 6 cm
outside radius r2 = 8 cm
thermal conductivity k = 0.5 W/m°C
inside temperature t1 = 430°C
outside temperature t2 = 30°C
to find out
Determine the heat loss per 1-m length of this insulation
solution
we know thermal resistance formula for cylinder that is express as
Rth = .................1
here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity
so
heat loss is change in temperature divide thermal resistance
Q =
Q =
Q = 4368.145 W
so heat loss per 1-m length of this insulation is 4368.145 W