An icebox made of 1-5 cm thick Styrofoam has dimensions 60 cm - 60 cm - 30 cm- It contains ice at 0- C and is kept in a room at 40- C- Find the rate at which the ice is melting- Latent heat of fusion of ice -3-36 - 105 J - kg- and thermal conductivity of Styrofoam -0-04 W - m - C- A- 0-316 g - sB- 0-46 g - sC- 1-46 g - s-D- 0-66 g - s

The correct option is B 0.46 g/sThe total surface area of the walls =2(60 cm × 60 cm + 60 cm × 30 cm + 60 cm × 30 cm) =1.44 m^2 The thickness of the walls = 1. ...

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Standard XII

Physics

Conduction Law: Differential Form

An icebox mad...

Question

An icebox made of 1.5 cm thick Styrofoam has dimensions $60 c m \times 60 c m \times 30 c m$ . It contains ice at $0^{\circ} C$ and is kept in a room at $40^{\circ} C$ . Find the rate at which the ice is melting. Latent heat of fusion of ice = $3.36 \times 10^{5} J / k g$ . and thermal conductivity of Styrofoam = $0.04 W / m^{\circ} C$ .

0.46 g/s

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1.46 g/s.

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0.66 g/s

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0.316 g/s

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Solution

The correct option is A 0.46 g/s
The total surface area of the walls
$= 2 (60 c m \times 60 c m + 60 c m \times 30 c m + 60 c m \times 30 c m)$
$= 1.44 m^{2}$
The thickness of the walls = 1.5 cm = 0.015 m.
The rate of heat flow into the box is
$\frac{Δ Q}{Δ t} = \frac{K A (θ_{1} - θ_{2})}{x} = \frac{(0.04 W / m - r C) (1.44 m^{2}) (40 r C)}{0.015 m} = 154 W$
The rate at which the ice melts is
$\frac{154 W}{3.36 \times 10^{5} J / k g} = 0.46 g / s$

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An icebox made of 1.5 cm thick Styrofoam has dimensions 60 cm×60 cm×30 cm. It contains ice at 0∘C and is kept in a room at 40∘C. Find the rate at which the ice is melting. Latent heat of fusion of ice = 3.36×105J/kg. and thermal conductivity of Styrofoam = 0.04W/m∘C.