Question

Figure (28-E4) 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 cm2. The junction is maintained at a constant temperature 40°C and the two ends are maintained at 80°C. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivites are KAt = 200 W m−1°C−1 and KCu = 400 W m−1°C−1.
Figure

Answer 1

Answer:

hey mate your answer is

Explanation:

Figure (28-E4) shows an aluminium rod joined to a copper rod. Each of the rods has a length of 20cm and area of cross-section 0.20cm2. The junction is maintained at a constant temperature 40°C and the two ends are maintained at 80°C. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivities are KAl = 200W/m-°C and KCu = 400W/m-°C. Read more on Sarthaks.com - https://www.sarthaks.com/63671/figure-28-e4-shows-an-aluminium-rod-joined-to-a-copper-rod-each-of-the-rods-has-length-of-20cm

Answer 2

Heat drawn is 144 Joules

Explanation:

Area of cross-section, $A = 0.20\ cm^2$

= $0.2 \times 10^-^4\ m^2$

Thermal conductivity of aluminium, $K_{Al} = 200\ Wm^-^1 C$

Thermal conductivity of copper, $K_{Cu} = 400\ Wm^-^1 C$

Total heat flowing per second = $q_{Al} + q_{Cu}$

= $\frac {K_{Al} \times A \times (80-40)}{l} + \frac {K_{Cu} \times A \times (80-40)}{l}$

= $\frac {200 \times 0.2 \times 10^-^4 \times 40}{0.2} + \frac {400 \times 0.2 \times 10^-^4 \times 40}{0.2}$

= $8 \times 10^-^1 + 16 \times 10^-^1$

= $24 \times 10^-^1$

= $24\ Jsec^-^1$

Heat drawn in 1 minute = 2.4 $\times$ 60 = 144 J