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Question

A refrigerator door is 150 cm high, 80 cm wide and 6 cm thick. If the coefficient of conductivity is 0.0005 cal/cm-s-C and the inner and outer surfaces are at 0C and 30C respectively, find the heat loss per minute through the door. (in Calories)

A
3600
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B
1800
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C
900
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D
1200
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Solution

The correct option is B 1800
Given,
Height of refrigerator (h)=150 cm
Width of refrigerator (w)=80 cm
Thickness of refrigerator (d)=6 cm
Coefficient of thermal conductivity (k)=0.0005 cal/cm-s-C=0.0005×60 cal/cm-min-C
Temperature difference between inner and outer surfaces (thtc)=30C

Applying the equation of thermal conductivity,
Qt=kA(thtc)d
For t=1 minute:
Q=0.0005×(60)×(150×80)×(30)6Q=1800 cal/min

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