Answer:
[tex]Q = 28 \times 10^{-3} J[/tex]
Explanation:
Heat transfer due to conduction is given by the formula
[tex]\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}[/tex]
here we know that
A = area of crossection
L = length through which heat will flow
[tex]\Delta T [/tex] = temperature difference across the ends of rod
K = thermal conductivity = 35 W/m K
now from above formula we can say
[tex]\frac{dQ}{dt} = \frac{35(13\times 10^{-4})(8 )}{13 \times 10^{-2}}[/tex]
[tex]\frac{dQ}{1 s} = 0.028[/tex]
[tex]Q = 28 \times 10^{-3} J[/tex]
