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Heat and Thermodynamics

Section: Physics

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Question : 77 of 78

Marks: +1, -0

The length of a metal bar is

20\ \mathrm{cm}

and the area of cross section is

4 \times 10^{-4}\ \mathrm{m}^2

. If one end of the rod is kept in ice at

0^{\circ}\ \mathrm{C}

and the other end is kept in steam at

100^{\circ}\ \mathrm{C}

, the mass of ice melted in one minute is

5\ \mathrm{g}

. The thermal conductivity of the metal in

\mathrm{Wm}^{-1}\ \mathrm{K}^{-1}

(Latent heat of fusion

=80\ \mathrm{cal} / \mathrm{gm}

[AP EAPCET 18 May 2024 Shift 1]

140
120
100
160

Solution:

Length of the rod,

x=20\ \mathrm{cm}=0.2\ \mathrm{m}

Temperatures at the ends:

\;T_1=0^{\circ}\ \mathrm{C}=273\ \mathrm{K}

\;T_2=100^{\circ}\ \mathrm{C}=373\ \mathrm{K}

Cross-sectional area,

A=4 \times 10^{-4}\ \mathrm{m}^2

Mass of ice melted,

m=5\ \mathrm{g}

Time,

t=1\ \mathrm{min}=60\ \mathrm{s}

Latent heat of fusion,

L_{\;\text{ice}\;}=80\ \mathrm{cal} / \mathrm{g} =336\ \mathrm{J} / \mathrm{g}

Calculation:

Total Heat Required to Melt the Ice:

Q=m \times L_{\;\text{ice}\;}=5 \times 336=1680\ \mathrm{J}

Rate of Heat Transfer:

\;\frac{Q}{t}=\;\frac{1680}{60}=28\ \mathrm{J} / \mathrm{s}

Thermal Conductivity (

K

):

\;K=\;\frac{\Delta Q \cdot x}{A \cdot \Delta T}=\;\frac{28 \times 0.2}{4 \times 10^{-4}(373-273)}

\;K=\;\frac{1.4 \times 10^4}{100}

\;K=1.4 \times 10^2

\;K=140\ \mathrm{W} / \mathrm{mK}

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