Question

a lower surface of a cuboidal slab of stone of side 0.1 is exposed to steam at 373 Kelvin and thick layer of ice covers the upper surface and the other faces are covered by non conducting material in 40 minutes 0.15 kg of ice melts. find the thermal conductivity for the stone
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Answer 1

Answer:

Thermal conductivity of the stone is 2.1 W/m K

Explanation:

As we know by the law of thermal conduction we have

$\frac{dQ}{dt} = \frac{KA}{L}(\Delta T)$

so here we know that due to heat transfer 0.15 kg ice melt in 40 minutes

so we have

$\frac{dQ}{dt} = \frac{mL}{t}$

$\frac{dQ}{dt} = \frac{0.15 \times 336000}{40 \times 60}$

$\frac{dQ}{dt} = 21 J/s$

now we have

$21 = \frac{K(0.1 \times 0.1)(373 - 273)}{0.1}$

$21 = (K\times 0.1 \times 100)$

$K = 2.1 W/m K$

#Learn

Topic : Thermal conduction

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