Question

the lower surfacr of cubical slab of a stone of side 0.1m is exposed to steam at 373k and thick layer of ice covers tge upper surface and other faces are covered by non conducting material in 40 min 0.15kg of ice melts .find thermal conductivity of a stone ? latent heat =336×10​

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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