First slide

Conduction

Question

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 at 0°C rests on the upper surface of the slab. In 56 minutes 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)

Moderate

A

2.05 J/m/s/°C
B

1.02 J/m/s/°C
C

1.33 J/m/s/°C
D

1.29 J/m/s/°C

Solution

$\frac{∆ Q}{∆ t} = KA \frac{∆ T}{∆ x} = \frac{mL}{∆ t} \Rightarrow K = \frac{mL ∆ x}{A ∆ T ∆ t}$
$K = \frac{4.8 \times 3.36 \times 10^{5} \times 0.1}{56 \times 60 \times 100 \times 0.36} = 1.33 J / m / s /^{o} C$

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