Question

A cubical thermocole ice box of side 20 cm has a thickness 5 cm. If 5 kg of ice is put in the box, the amount of ice remaining after 10 hours is

(The outside temperature is 50°C and the coefficient of thermal conductivity of thermocole

$= 0.01Js^{- 1}{m^{- 1}}^{o}C^{- 1}$latent heat of fusion of ice

$= 335 \times 10^{3}Jkg^{- 1}$)

• A. 3.7 kg
• B. 3.9 kg
• C. 4.7 kg
• D. 4.9 kg

Answer 1

Answer: (C)

4.7 kg

Answer 2

Here, Length of each side,

l = 20 cm = 20 × $10^{- 2}m$

Thickness, x = 5cm = 5 × $10^{- 2}m$

Total surface area through which heat enters into the box, $A = 6l^{2} = 6 \times (20 \times 10^{- 2}m)^{2} = 24 \times 10^{- 2}m^{2}$

Temperature difference, $\Delta T = 50{^\circ}C - 0{^\circ}C = 50{^\circ}C`$

Time, t = 10 h = 10 × 60 × 60s = 36 × $10^{3}s$

$L_{f} = 335 \times 10^{3}Jkg^{- 1}$

Total heat entering the box through all the six faces is

$Q = \frac{KA\Delta Tt}{x}$

$= \frac{0.01Js^{- 1}m^{- 1}{^\circ}C^{- 1} \times 24 \times 10^{- 2}m^{2} \times 50{^\circ}C \times 36 \times 10^{3}s}{5 \times 10^{- 2}m}$= 86400 J ….. (i)

Let m kg of ice melts in this time

$\therefore Q = mL_{f}$ …… (ii)

From (i) and (ii), we get

m = $\frac{86400J}{335 \times 10^{3}Jkg^{- 1}} = 0.258kg$

$\therefore$Amount of ice left

$= (5 - 0.258)kg$

$= 4.724kg = 4.7kg$