A slab of stone are 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 100-C- A block of ice of 0-C rests on upper surface of the slab- In one hour 4-8 kg of ice is melted- The thermal conductivity of the stone in Js-1 m-1 k-1 is Latent heat of ice - 3-36- 105J-kg

The correct option is D 1.24 Qt = KA(Q_1Q_2)l = 48× 3.36× 10^560× 60 = K× 3600× 10^ 4× 10010× 10^ 2 = 1.24 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

1st Law of Thermodynamics

A slab of sto...

Question

A slab of stone are $3600 c m^{2}$ and thickness $10 c m$ is exposed on the lower surface to steam at $100^{\circ} C$ . A block of ice of $0^{\circ} C$ rests on upper surface of the slab. In one hour $4.8 k g$ of ice is melted. The thermal conductivity of the stone in $J s^{- 1} m^{- 1} k^{- 1}$ is (Latent heat of ice $= 3.36 \times 10^{5} J / k g$ )

12.0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.02

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.24

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $1.24$
$\frac{Q}{t} = \frac{K A (Q_{1} Q_{2})}{l} = \frac{48 \times 3.36 \times 10^{5}}{60 \times 60} = \frac{K \times 3600 \times 10^{- 4} \times 100}{10 \times 10^{- 2}} = 1.24$

Suggest Corrections

Similar questions

Q. 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

)

Q. A slab of stone area 3600 cm

^{2}

and thickness 10 cm is exposed on the lower surface to steam at 100

^{0} C

. A block of ice at

0^{0} C

rest on upper surface of the slab. In one minute, 4.8 kg of ice is melted. The thermal conductivity of the stone in

J s^{- 1} m^{- 1} k^{- 1}

is (latent heat of ice =

3.36 \times 10^{5}

J/kg)

Q. A slab of stone of area

0.36 m^{2}

and thickness

0.1

is exposed on the lower surface to steam at

100^{o} C

. A block of ice at

0^{o} C

rests on the upper surface of the slab. In one hour

4.8 k g

of ice is melted. The thermal conductivity of slab is :(Given latent heat of fusion of ice

= 3.36 \times 10^{5} J {k g}^{- 1}

)

A slab of stone of area $0.34 m^{2}$ and thickness 10 cm is exposed on the lower face to steam at $100^{\circ} C$ . A block of ice at $0^{\circ} C$ rests on the upper face of the slab. In one hour, 3.6 kg of ice is melted. Assume that the heat loss from the sides is negligible. The latent heat of fusion of ice is $3.4 \times 10^{4} J k g^{- 1}$ . What is the thermal conductivity of the stone in units of $J s^{- 1} m^{- 1} C^{- 1}$ ?

Join BYJU'S Learning Program

A slab of stone are 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 100∘C. A block of ice of 0∘C rests on upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of the stone in Js−1m−1k−1 is (Latent heat of ice =3.36×105J/kg)

The correct option is D 1.24Qt=KA(Q1Q2)l=48×3.36×10560×60=K×3600×10−4×10010×10−2=1.24

The correct option is D $1.24$
$\frac{Q}{t} = \frac{K A (Q_{1} Q_{2})}{l} = \frac{48 \times 3.36 \times 10^{5}}{60 \times 60} = \frac{K \times 3600 \times 10^{- 4} \times 100}{10 \times 10^{- 2}} = 1.24$