An ice box used for keeping eatable cold has a total wall area of 1 metre2 and a wall thickness of 5-0 cm- The thermal conductivity of the ice box is K-0-01 joule-metre-C- It is filled with ice at 0- along with eatables on a day when the temperature is 30- The latent heat of fusion of ice is 334- 103joules-kg - The amount of ice melted in one day is1 day -86-400 seconds

DQ/dt=KA/ldθ =0.01× 1/0.05× 30=6J/sec Heat transferred in on day (86400 sec) Q =6× 86400=518400J Now Q=mL⇒ m=Q/L=518400/334× 10^3 = 1.552 kg = 1552g. ...

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Standard XIII

Physics

Conduction Law

An ice box us...

Question

An ice box used for keeping eatable cold has a total wall area of $1 m e t r e^{2}$ and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is $K = 0.01 j o u l e / m e t r e -^{\circ} C$ . It is filled with ice at $0^{\circ}$ along with eatables on a day when the temperature is $30^{\circ}$ . The latent heat of fusion of ice is $334 \times 10^{3} j o u l e s / k g$ . The amount of ice melted in one day is
(1 day =86,400 seconds )

776 gms

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7760 gms

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11520 gms

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1552 gms

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Solution

The correct option is D
1552 gms

$\frac{d Q}{d t} = \frac{K A}{l} d θ = \frac{0.01 \times 1}{0.05} \times 30 = 6 J / s e c$
Heat transferred in on day (86400 sec)
$Q = 6 \times 86400 = 518400 J$
Now $Q = m L \Rightarrow m = \frac{Q}{L} = \frac{518400}{334 \times 10^{3}}$
$= 1.552 k g = 1552 g$ .

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Similar questions

An ice box used for keeping eatable cold has a total wall area of $1 m e t r e^{2}$ and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is $K = 0.01 j o u l e / m e t r e -^{\circ} C$ . It is filled with ice at $0^{\circ}$ along with eatables on a day when the temperature is $30^{\circ}$ . The latent heat of fusion of ice is $334 \times 10^{3} j o u l e s / k g$ . The amount of ice melted in one day is
(1 day =86,400 seconds )

Q. An ice box used for keeping eatables cool has a total wall area of

1 m^{2}

and a wall thickness of

5.0 c m

. The thermal conductivity of the ice box is

K = 0.01 J / m^{o} C

. It is filled with ice along with eatables on a day when the temperature is

30^{o} C

. The latent heat of fusion of ice is

334 \times 10^{2} J / k g

. The amount of ice melted in one day is (

1

day

= 86, 400

seconds):

An ice box used for keeping eatable cold has a total wall area of 1 m² and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is K = 0.01 joule/metre-⁰C. It is filled with ice at ⁰C along with eatables on a day when the temperature is 30⁰C. The latent heat of fusion of ice is 334×10³ joule/kg. The amount of ice melted in one day is ( 1day = 86.400 s)

Q. A thermocol box has a total wall area (including the lid) of

1.0 m^{2}

and wall thickness of

3 c m

. It is filled with ice at

0^{o} C

. If the average temperature outside the box is

30^{o} C

throughout the day, the amount of ice that melts in one day is
(Use

K_{t h e r m o c o l} = 0.03 W / m K, L_{f u s i o n (i c e)} = 3.00 \times 10^{5} J / k g

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The correct option is D 1552 gms dQdt=KAldθ=0.01×10.05×30=6J/sec Heat transferred in on day (86400 sec) Q=6×86400=518400J Now Q=mL⇒m=QL=518400334×103 =1.552kg=1552g.

The correct option is D
1552 gms

$\frac{d Q}{d t} = \frac{K A}{l} d θ = \frac{0.01 \times 1}{0.05} \times 30 = 6 J / s e c$
Heat transferred in on day (86400 sec)
$Q = 6 \times 86400 = 518400 J$
Now $Q = m L \Rightarrow m = \frac{Q}{L} = \frac{518400}{334 \times 10^{3}}$
$= 1.552 k g = 1552 g$ .