A thermocole cubical ice box of side 30 cm is of thickness 5 cm. If 4kg of ice is put in the box, estimate the amount of ice remaining after 6 hours. The outside temperature is 45 0 c and coefficient of thermal conductivity of thermocole = 0.01w/mk.
Answers
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 × 103 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 = 6s2 = 6 × (0.3)2 = 0.54 m3
θ = 0.01 × 0.54 × 45 × 6 × 60 × 60 / 0.05 = 104976 J
But θ = m’L
∴ m‘ = θ/L
= 104976/(335 × 103) = 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.
Answer:
A cubical thermocol ice box of side length
30 Cm
has thickess of
5.2 Cm
If
4.0 kg
of ice is put in the box estimate the amount of ice remaining after
6 hr
The outside temperature is
45° C
and co-efficient of the thermal conductivity of the thermocol is
0.01 Js-1m-1 k 1
(Latent Heat of fusion of water
=335×10cube J kg -1