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Byju's Answer

Standard XII

Physics

Conduction Law

A slab of sto...

Question

A slab of stone area $3500 {c m}^{2}$ and thickness $10 c m$ is exposed on the lower surface to steam at $100^{o} C$ . A block of ice at $0^{o} C$ rests on upper surface of the slab. In one hour $4.8 k g$ of ice of melted. The thermal conductivity of the stone is $J s^{- 1}$ $m^{- 1}$ $k_{- 1}$ is
(Latent heat of ice $= 3.36 \times 10^{5} J / k g$ )

12.0

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10.5

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1.02

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1.24

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Solution

The correct option is B $1.24$
Given : $A = 3500 c m^{2} = 0.35 m^{2}$ $l = 10 c m = 0.1 m$ $Δ T = 100 - 0 = 100^{o} C$
Mass of ice melted $m = 4.8 k g$
Time taken $t = 1 h r = 3600 s$
Latent heat of ice $L = 3.36 \times 10^{5} J / k g$
Heat absorbed by ice = Heat conducted by slab
$∴$ $m L = \frac{K A t Δ T}{l}$
Or $4.8 \times 3.36 \times 10^{5} = \frac{K (0.35) (3600) (100)}{0.1}$
$⟹ K = 1.24 J s^{- 1} m^{- 2} k^{- 1}$

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A slab of stone area 3500cm2 and thickness 10cm is exposed on the lower surface to steam at 100oC. A block of ice at 0oC rests on upper surface of the slab. In one hour 4.8kg of ice of melted. The thermal conductivity of the stone is Js−1 m−1 k−1 is(Latent heat of ice =3.36×105J/kg)

The correct option is B 1.24Given : A=3500 cm2=0.35 m2 l=10 cm=0.1 m ΔT=100−0=100oCMass of ice melted m=4.8 kgTime taken t=1 hr=3600 sLatent heat of ice L=3.36×105 J/kgHeat absorbed by ice = Heat conducted by slab∴ mL=KAtΔTlOr 4.8×3.36×105=K(0.35)(3600)(100)0.1⟹ K=1.24 Js−1m−2k−1