Transmission Of Heat Question 124
Question: An ice box used for keeping eatable cold has a total wall area of $ 1\ metre^{2} $ and a wall thickness of $ 5.0cm $ . The thermal conductivity of the ice box is $ K=0.01\ joule/metre{{-}^{o}}C $ . It is filled with ice at $ 0^{o}C $ along with eatables on a day when the temperature is 30°C. The latent heat of fusion of ice is $ 334\times 10^{3}joules/kg $ . The amount of ice melted in one day is ($ 1day=86,400\ \sec onds $ )
[MP PMT 1995]
Options:
A) $ 776\ gms $
B) $ 7760\ gms $
C) $ 11520\ gms $
D) $ 1552\ gms $
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Answer:
Correct Answer: D
Solution:
$ \frac{dQ}{dt}=\frac{KA}{l}d\theta $
$ =\frac{0.01\times 1}{0.05}\times 30 $ = 6J/sec Heat transferred in on day (86400 sec) $ \theta =6\times 86400=518400,J $ Now $ Q=mL $
Therefore $ m=\frac{Q}{L} $
$ =\frac{518400}{334\times 10^{3}} $ = 1.552 kg = 1552g.
BETA
