__Question 11.19__:
A
‘thermacole’ icebox is a cheap and efficient method for
storing small quantities of cooked food in summer in particular. A
cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of
ice is put in the box, estimate the amount of ice remaining after 6
h. The outside temperature is 45 °C, and co-efficient of thermal
conductivity of thermacole is 0.01 J s

^{–1}m^{–1}K^{–1}. [Heat of fusion of water = 335 × 10^{3}J kg^{–1}]

__Solution__:*s*= 30 cm = 0.3 m

Thickness of the ice box,

*l*= 5.0 cm = 0.05 m

Mass of ice kept in the ice box,

*m*= 4 kg

Time gap,

*t*= 6 h = 6 × 60 × 60 s

Outside temperature,

*T*= 45°C

Coefficient of thermal conductivity of thermacole,

*K*= 0.01 J s

^{–1}m

^{–1}K

^{–1}

Heat of fusion of water,

*L*= 335 × 10

^{3}J kg

^{–1}

Let

*m*

^{’}be the total amount of ice that melts in 6 h.

The amount of heat lost by the food:

θ = KA(T - 0)t / l

Where,

*A*= Surface area of the box = 6

*s*

^{2}= 6 × (0.3)

^{2}= 0.54 m

^{3}

^{}θ = 0.01 X 0.54 X45 X 6 X 60 X 60 / 0.05 = 104976 J

But θ = m'L

∴ m' = θ/L

= 104976/(335 X 10

^{3}) = 0.313 kg

Mass of ice left = 4 – 0.313 = 3.687 kg

Hence, the amount of ice remaining after 6 h is 3.687 kg.

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