### Q.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}]

**Answer:-**

Side of the given cubical ice box, *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 × 0.54 × 45 × 6 × 60 × 60 / 0.05 = 104976 J

But θ = m’*L*

∴ *m*‘ = θ/*L*

= 104976/(335 × 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.