Question

The three rods shown in figure have identical
dimensions. Heat flows from the hot end at a rate
of 40 W in the arrangement (a). Find the rates of
heat flow when the rods are joined as in
arrangement (b). (Assume K. = 200 W/m "C and
K = 400 W/m "C)​

Answer 1

Answer:

Dear friend, the image is not attached with the question....finding the answers would be a tough task....✌

Answer 2

Given that,

Conductivity of K $K_{Al}=200\ W/m$

Conductivity of Cu $K_{Cu}=400\ W/m$

Rate = 40 W

According to figure,

Conductivity in series in fig a

We need to calculate the net conductivity

Using formula of series

$\dfrac{1}{K}=\dfrac{1}{K_{Al}}+\dfrac{1}{K_{Cu}}+\dfrac{1}{K_{Al}}$

$\dfrac{1}{K}=\dfrac{2}{K_{Al}}+\dfrac{1}{K_{Cu}}$

Put the value into the formula

$\dfrac{1}{K}=\dfrac{2}{200}+\dfrac{1}{400}$

$K=80\ W/m k$

In series, heat flows at a rate of $I_{1}=40\ W$

(b). Conductivity in parallel in fig a

We need to calculate the net conductivity

Using formula of parallel

$K'=K_{Al}+K_{Cu}+K_{Al}$

Put the value into the formula

$K'=200+400+200$

$K'=800\ W/m k$

We know that,

The ratio of current for constant temperature difference

We need to calculate the rate of heat flows from the hot end

Using the formula of rate of heat

$\dfrac{I_{1}}{I_{2}}=\dfrac{K}{K'}$

$I_{2}=\dfrac{K'}{K}\times I_{1}$

Put the value into the formula

$I_{2}=\dfrac{800}{80}\times 40$

$I_{2}=400\ W$

Hence, The Heat flows from the hot end at a rate  of 400 W.