Question

A cubical ice box of thermocole has each side equal to 30 cm and a
thickness of 5 cm. 4 kg of ice is put in the box. If outside temperature is
45 °C, coefficient of thermal conductivity is 0.01 J/s/m/°C, and latent
heat of fusion of ice is 335 x 10 J/kg, then the mass of ice left after 6
hours is
(a) 0.313 kg
(b) 3.687 kg
(c) 36.87 kg
(d) 31.3 kg
​

Answer 1

SOLUTION:

Given:

• A = 6 * side²

A = 6 * 30 * 30

A = 5400 cm sq.

A = 0.54 m sq.

• x = 5 cm = 0.05 m
• t = 6 h = 6*3600 sec

$T_{1} - T_{2} = 45 - 0 = 45^{o}C$

$K = 0.01 Js^{-1}m^{-1}^{o}C^{-1}$

$L = 335 \times 10^{3} jkg^{-1}$

The total heat entering the box through all the six faces,

$Q = \dfrac{KA(T_{1} -T_{2})t}{x}$

$Q = \dfrac{0.01 \times 0.54 \times 45 \times 6 \times 6300}{0.05}$

$\boxed{Q = 104976 J}$

Let (m) kg of ice melt due to this heat.

Then

Q = mL

OR

$m = \frac{Q}{L} = \frac{104976J}{336 \times 10^{3}Jkg^{-1}} = 0.313 Kg$

• Mass of ice left after six hours = 4 - 0.313 = 3.687 kg

Hence,

Option (b) 3.687 kg is the correct answer