Question

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 × 103 J kg–1]
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Answer 1

Solution:

Initial temperature of the body of the child, T1 = 101°F

Final temperature of the body of the child, T2 = 98°F

Change in temperature, ΔT = [ (101 - 98) × (5/9) ] o C

Time taken to reduce the temperature, t = 20 min

Mass of the child, m = 30 kg = 30 × 103 g

Specific heat of the human body = Specific heat of water = c

= 1000 cal/kg/ °C

Latent heat of evaporation of water, L = 580 cal g–1

The heat lost by the child is given as:

∆θ = mc∆T

= 30 × 1000 × (101 - 98) × (5/9)

= 50000 cal

Let m1 be the mass of the water evaporated from the child’s body in 20 min.

Loss of heat through water is given by:

∆θ = m1L

∴ m1 = ∆θ / L

= (50000 / 580) = 86.2 g

∴ Average rate of extra evaporation caused by the drug = m1 / t

= 86.2 / 200

= 4.3 g/min.

Answer 2

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Answer is in the attachment provided