An ice cube of dimensions 60 cm × 50 cm × 20 cm is placed in an insulation box of wall thickness 1 cm . The box keeping the ice cube at 0 ° C of temperature is brought to a room of temperature 40 ° C . The rate of melting of ice is approximately: (Latent heat of fusion of ice is 3 . 4 × 10 5 J kg - 1 and thermal conducting of insulation wall is 0 . 05 W m - 1 ° C - 1 )

Step-by-step solution

Using the equation of conduction, d Q d t = K A Δ T l Now the total area of the box will be, A = 2 0 . 6 × 0 . 5 + 0 . 5 × 0 . 2 + 0 . 2 × 0 . 6 A = 2 0 . 3 + 0 . 1 + 0 . 12 A = 2 0 . 52 = 1 . 04   m 2 Therefore, d Q d t = K A Δ T l = 0 . 05 × 1 . 04 × 40 0 . 01 = 208   J   s - 1 This heat will melt the ice. Therefore, d Q d t = d m d t L ⇒ 208 = d m d t × 3 . 4 × 10 5 ⇒ d m d t = 61 × 10 - 5   kg   s - 1