In figure shows an aluminium rod joined to a copper rod- Each of the rods has a length of 20 cm and area of cross section 0-20 cm 2- The junction is maintained at a constant temperature 40- and the two ends are maintained at 80- - Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached- The conductivities are K Al -200 W m -1- C -1 and K Cu -400 W m -1- C -1- 40 -circ C M80- circ C array - c-c - -line-endarray 80 - C -

K_Al = 200 W m^ 1 ∘ C^ 1 K_Cu = 400 W m^ 1 ∘ C^ 1 A = 0.2 cm^2 = 2 × 10^ 5 m^2 l = 20 cm =2 × 10^ 1 m Heat drawn per sec. Q_Al+ Q_Cu = K_Al× A × (80 40)/l+ K ...

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Standard XII

Physics

Conduction Law

In figure sho...

Question

In figure shows an aluminium rod joined to a copper rod. Each of the rods has a length of 20 cm and area of cross section $0.20 c m^{2}$ . The junction is maintained at a constant temperature $40^{\circ}$ and the two ends are maintained at $80^{\circ}$ . Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivities are $K_{A l} = 200 W m^{- 1 \circ} C^{- 1}$ and $K_{C u} = 400 W m^{- 1 \circ} C^{- 1}$

$40^{\circ} C 80^{\circ} C \begin{matrix} Aluminium & Copper \end{matrix} 80^{\circ} C$

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Solution

$K_{A l} = 200 W m^{- 1 \circ} C^{- 1}$

$K_{C u} = 400 W m^{- 1 \circ} C^{- 1}$

A = 0.2 $c m^{2} = 2 \times 10^{- 5} m^{2}$

l = 20 cm = $2 \times 10^{- 1} m$

Heat drawn per sec.

$Q_{A l} + Q_{C u} = \frac{K_{A l} \times A \times (80 - 40)}{l} + \frac{K_{C u} \times A \times (80 - 40)}{l}$

$= \frac{2 \times 10^{- 5} \times 40}{2 \times 10^{- 1}} [200 + 400]$

= 2.4 J

Heat drawn per minute = $2.4 \times 60 = 144 J$

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Q. Two rods

A

and

B

of different materials but same cross section are joined as shown in figure. The free end of

A

is maintained at

100^{\circ} C

and the free end of

B

is maintained at

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KAl=200 Wm−1∘C−1 KCu=400Wm−1∘C−1 A = 0.2 cm2=2×10−5m2 l = 20 cm =2×10−1m Heat drawn per sec. QAl+QCu=KAl×A×(80−40)l+KCu×A×(80−40)l =2×10−5×402×10−1[200+400] = 2.4 J Heat drawn per minute = 2.4×60=144J