Question

A metal cylinder of length 0.05m having its cross section area and ends of cylinder are kept at 3450K and 2850K . Calculate the amount of heat transmitted in 20 minutes from one end at the higher temperature to the other end at lower temperature of the cylinder .
given(the termal conductivity if the metal cylinder is 419wm-1 k-1)
$(wrong \: answer \:report \: kardenge)$
​

Answer 1

Explanation:

hope it is very helpful to you

pls mark as brainlist answer

Answer 2

Answer:

given:

• A=0.9m²
• θ1=3450k
• θ2=2850k
• d=0.05m
• min=1200s

formula:

$Q = \frac{kA(θ_1 -θ_2)t }{d}$

$\frac{419 \times 0.9(3450 - 2850) \times 1200}{0.05}$

$q = 5.43 \times 10^9j$

hope it helps....