A rod of negligible heat capacity has length 20 cm- area of cross section 1-0 cm 2 and thermal conductivity 200 W m -1- C -1- The temperature of one end is maintained at 0- C and that of the other end is slowly andlineraly varied from 0- C to 60- C in 10 minutes- Assuming no loss of heat through the sides- find the total heat transmitted throug the rod in these 10 minutes-

D θ/dt = 60/10 × 60 = 0.10 ^∘C /sec d Q/dt= KA/d(θ_2 θ _1) = KA0.1 /d + KA0.2 /d+ ....+ KA60 /d = KA/d(0.1 +0.2+ ....+60) = KA/d600/2 (2 × 0.1 +599 × 0.1) [ a ...

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Standard XII

Physics

Conduction Law

A rod of negl...

Question

A rod of negligible heat capacity has length 20 cm, area of cross section 1.0 $c m^{2}$ and thermal conductivity $200 W m^{- 1 \circ} C^{- 1}$ . The temperature of one end is maintained at $0^{\circ} C$ and that of the other end is slowly andlineraly varied from $0^{\circ} C$ to $60^{\circ} C$ in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted throug the rod in these 10 minutes.

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Solution

$\frac{d θ}{d t} = \frac{60}{10 \times 60} = {0.10}^{\circ} C / s e c \frac{d Q}{d t} = \frac{K A}{d} (θ_{2} - θ_{1})$

$= \frac{K A 0.1}{d} + \frac{K A 0.2}{d} + . . . . + \frac{K A 60}{d}$

$= \frac{K A}{d} (0.1 + 0.2 + . . . . + 60)$

$= \frac{K A}{d} \frac{600}{2} (2 \times 0.1 + 599 \times 0.1)$

$[a + 2 a + . . . n a = \frac{n}{2} 2 a + (n - 1) a]$

$= \frac{200 \times 1 \times 10^{- 4}}{20 \times 10^{- 2}} \times 300 \times (0.2 + 59.9)$

$= \frac{200 \times 10^{- 2} \times 300 (60.1)}{20}$

$= 3 \times 10 \times 60.1$

= 1803 W = 1800 W

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dθdt=6010×60=0.10∘C/secdQdt=KAd(θ2−θ1) =KA0.1d+KA0.2d+....+KA60d =KAd(0.1+0.2+....+60) =KAd6002(2×0.1+599×0.1) [a+2a+...na=n22a+(n−1)a] =200×1×10−420×10−2×300×(0.2+59.9) =200×10−2×300(60.1)20 =3×10×60.1 = 1803 W = 1800 W