Question

Sunlight of intensity 0.6 kW m falls on a patch of ice. Assuming the ice absorbs
20% of the light; calculate what thickness of ice would melt in 1 minute. (Assume any
water produced runs off)​

Answer 1

answer : 2.3446 × 10^-3 cm

intensity of sunlight , I = 0.6kW/m²

so, power of sunlight per m² , P = 0.6kW = 600 watt.

energy radiated by sunlight in 1 minute , E = power × 1 minute

= 600 watt × 60sec

= 36000 J = 3.6 ×10⁴ J

a/c to question, ice absorbs 20% of the light.

so, amount of heat energy gain by ice = 20% of 3.6 × 10⁴ J

= 20 × 3.6 × 10⁴/100

= 7.2 × 10³ J

now, heat gained by ice = mLv

m is mass of ice is melted and Lv is latent heat of fusion of ice e.g., Lv = 334 × 10³ J/Kg

now, 7.2 × 10³ J = m × 334 × 10³ J/Kg

or, 7.2/334 kg = m

or, m = 21.5g

now, mass of ice is melted = = 21.5 g

we know, density of ice at 0°C is 0.917g/cm³

so, volume of water = 21.5g/(0.917g/cm³) = 23.446 cm³

cross sectional area of ice block is 1m²

and Let us consider that thickness of ice block is t

then, volume of ice = 1m² × t

or, 23.446 × 10^-6 m³ =1 m² × t

or, t = 23.446 × 10^-6 m or, 23.446 × 10^-4 cm or 2.3446 × 10^-3 cm