An ice box used for keeping eatable cold has a total wall area of 1 metre-2 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- 10-3joules-kg - The amount of ice melted in one day is -1 day -86-400 seconds - 776 gms7760 gms1552 gms11520 gms

The correct option is C 1552 gmsdQ/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× 1 ...

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Byju's Answer

Standard XII

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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The correct option is D 1552 gmsdQdt=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 $Q = m L \Rightarrow m = \frac{Q}{L} = \frac{518400}{334 \times 10^{3}}$
$= 1.552 k g = 1552 g$ .