A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100 °C. A block of ice 0 °C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is:

(Given latent heat of fusion of ice = 3.36 x 105 J kg-1)

  • Option 1)

    1.29 J/m/s/°C

  • Option 2)

    2.05 J/m/s/°C

  • Option 3)

    1.02 J/m/s/°C

  • Option 4)

    1.24 J/m/s/°C

 

Answers (1)

In one hour 4.8 kg of ice is muted 

\therefore amount of heat transferred from hot end to cold end 

=mL=4.8\times 3.36\pi 10^{5}J

Q=16.128\times 10^{5}J in one hour 

Rate of transfer of heat 

\frac{\Delta Q}{\Delta t}=\frac{16.128\times 10^{5}}{3600}=4.48\times 10^{2}J/sec

Also \frac{\Delta Q}{\Delta t}=\frac{KA-\Delta \theta }{L}

\frac{K\times 0.36\times 100}{0.1}=448\\ K=\frac{44.8}{36}=1.24


Option 1)

1.29 J/m/s/°C

This option is incorrect 

Option 2)

2.05 J/m/s/°C

This option is incorrect 

Option 3)

1.02 J/m/s/°C

This option is incorrect 

Option 4)

1.24 J/m/s/°C

This option is correct 

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