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`        A thermocol ice box is cheap and efficient method for storing small quantities of cooked food in a summer particular. a cubical ice box of side side 30 cm has a thickness of 5 cm if 4 kg of ice is put  in the box estimate the amount of ice remaining after 6h . the outside temperature is 45°C and coefficient of thermal conductivity of thermocol is 0.01j(smk)^-1  heat of fusion of water is 355 *10³jkg^-1 `
one year ago

```							Side of the given cubical ice box, s = 30 cm = 0.3 mThickness of the ice box, l = 5.0 cm = 0.05 mMass of ice kept in the ice box, m = 4 kgTime gap, t = 6 h = 6 × 60 × 60 sOutside temperature, T = 45°CCoefficient of thermal conductivity of thermoCol, K = 0.01 J s–1 m–1 K–1Heat of fusion of water, L = 335 × 103 J kg–1Let m’ = total amount of ice that melts in 6 h.The amount of heat lost by the food:θ = (KA (T1 – T2)/t)/d  Where,A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3θ = (0.01 x 0.54 x (45) x 6 x 60 x 60)/ (0.05)=104976 JBut θ = m’LOr m’ = (θ /L)= (104976)/ (335 x 103)=0.313KgMass of ice left = (4 – 0.313) = 3.687 kgHence, the amount of ice remaining after 6 h is 3.687 kg.
```
one year ago
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