Thermal Properties of Matter
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Question : 19
Total: 22
A ‘thermocole’ 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 co-efficient of thermal conductivity of thermocole is 0.01 J s – 1 m – 1 K – 1
[Heat of fusion of water= 335 × 10 3 J k g – 1 ]
[Heat of fusion of water
Solution:
Here, length of each side, l = 30 c m = 0.3 m
Thickness of each side,Δ x = 5 c m = 0.05 m
total surface area through which heat enters into the box,
A = 6 l 2
= 6 × 0.3 × 0.3 = 0.54 m 2
Temp. diff.,Δ T = 45 – 0 = 45 ° C
K = 0.01 J s − 1 m − 1 K − 1
time,Δ t = 6 h r s = 6 × 60 × 60 s
Latent heat of fusion,L = 335 × 10 3 J k g − 1
Letm
∴ Δ Q = m L = K A (
) Δ t
m = K A (
)
= 0.01 × 0.54 ×
×
= 0.313 k g
Mass of ice left= 4 − .313 = 3.687 kg
Thickness of each side,
total surface area through which heat enters into the box,
Temp. diff.,
time,
Latent heat of fusion,
Let
Mass of ice left
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