An ice box used for keeping eatables cool has a total wall area of 1 m 2 and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is K=0.01 J / m ° C. It is filled with large amount of ice at 0° C along with eatables on a day when the temperature is 30° C. The latent heat of fusion of ice is 334 × 103 J / kg. The amount of ice melted in one day is ( 1 day =86,400 s )

Question

An ice box used for keeping eatables cool has a total wall area of 1 m 2 and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is K=0.01 J / m ° C. It is filled with large amount of ice at 0° C along with eatables on a day when the temperature is 30° C. The latent heat of fusion of ice is 334 × 103 J / kg. The amount of ice melted in one day is ( 1 day =86,400 s ) (A) 776 g (B) 7760 g (C) 11520 g (D) 1552 g

Tardigrade Admin · Accepted Answer

Quantity of heat transferred through wall will be utilized in melting of ice.

Q = \frac{K A Δ θt}{Δ x} = m L

∴

Amount of ice melted

m = \frac{K A Δ θt}{Δ xL}

∴ m = \frac{0.01 \times 1 \times ( 30 - 0 ) \times 86400}{5 \times 1 0 ^{- 2} \times 334 \times 1 0 ^{3}}

= 1.552 k g

or

1552 g m

$776 g$

$7760 g$

$11520 g$

$1552 g$

776g

7760g

11520g

1552g

Quantity of heat transferred through wall will be utilized in melting of ice. Q=ΔxKAΔθt​=mL ∴ Amount of ice melted m=ΔxLKAΔθt​ ∴m=5×10−2×334×1030.01×1×(30−0)×86400​ =1.552kg or 1552gm

$776 g$

$7760 g$

$11520 g$

$1552 g$