Question

A metal rod of length 10 cm and diameter 2 cm is covered with a non conducting substance. One end of it is maintained at 80°C, while the other end is put at 0°C. It is found that 20 g of ice melts in 5 min. Find the coefficient of thermal conductivity of metal in $Js^{- 1}m^{- 1}K^{- 1}$ (Take latent heat of ice is 80 cal $g^{- 1}$)

• A. 4.2
• B. 6.8
• C. 7.2
• D. 8.9

Answer 1

Answer: (D)

8.9

Answer 2

Here, length of the rod, $\Delta x = 10$cm = $10 \times 10^{- 2}m$ Radius of the rod $= \frac{2}{2} = 1cm = 10^{- 2}m$

Area of cross sections,

$A = \pi r^{2} = \pi(10^{- 2}) = \pi \times 10^{- 4}m^{2}$Now,

$\Delta T = 80 - 0 = 80{^\circ}C$

Mass of ice melted,

$m = 20g,L = 80calg^{- 1}$ Heat conducted,

$\Delta Q = mL = 20 \times 80 = 1600cal = 1600 \times 1.2J$

And $\Delta T = 5\min = 5 \times 60 = 300s$As $\frac{\Delta Q}{\Delta t} = KA\frac{\Delta T}{\Delta x}$

Or $K = \frac{\Delta Q/\Delta t}{A\Delta T/\Delta x} = \frac{\Delta Q\Delta x}{\Delta tA\Delta T}$

$= \frac{1600 \times 4.2 \times 10 \times 10^{- 2}}{300 \times \pi \times 10^{- 4} \times 80} = 8.9Js^{- 1}m^{- 1}K^{- 1}a$