Question

QUESTION :

A ' thermacole ' icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box , estimate the amount of ice remaining after 6 h. The outside temperature is 45° c and coefficient of thermal conductivity of thermacole is 0.01 $J{s}^{-1}\:{m}^{-1}{}^{\degree}\:{c}^{-1}$.
[ Heat of fusion of water = 335 x 10³ J${kg}^{-1}]$

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Answer 1

Answer:

Hence, the amount of ice remaining after 6 h is 3.687 kg.

Explanation:

Here,

side length of cubical ice-box = 30cm

area of 6 face of box ( A) = 6 × ( side)²

= 6 × (30)² = 5400 cm²

= 0.54 m²

Thickness of the box ( d) = 5cm = 5 × 10^-2 m

Mass of the ice ( m) = 4 kg

Time ( t) = 6h = 21600 sec

Difference in temperature ( ∆T) = final temperature - initial temperature

= 45°C - 0°C = 45°C

Latent heat of fusion of water ( Lf)= 335 × 10³ j/kg

Coefficient of thermal conductivity ( K) = 0.01 J/s.m.K

Let 'm' is the mass of ice is melted .

Heat supplied by the surrounding= heat taken by ice during melting

KA∆T.t /d = m' Lf

m' = KA∆T.t/Lf.d

= 0.01 × 0.54 × 45 × 21600/335 × 10³×5×10^-2

= 0.313 kg

Hence, mass remains in the box = m-m'

= 4 kg - 0.313 kg

= 3.687 Kg

hope it helps u ✌

Answer 2

Answer:

the amount of ice remaining after 6 h is 3.687 kg

Explanation:

Side of the given cubical ice box, s=30cm=0.3m

Thickness of the ice box, l=5.0cm=0.05m

Mass of ice kept in the ice box, m=4kg

Time gap, t=6h=6×60×60s

Outside temperature, T=45°C

Coefficient of thermal conductivity of thermacole, K=0.01Js

Heat of fusion of water, L=335×10

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 =6s^2

=6×(0.3) s

=0.54m^3

θ=0.01×0.54×45×6×60/0.05=104976J

But θ=m /l

∴m =θ/L

=104976/(335×10^3 )=0.313kg

Mass of ice left =4–0.313=3.687kg

Hence,the amount of ice remaining after 6 h is 3.687 kg.