A refrigerator door is 150cm high, 80cm wide and 6cm thick. If the coefficient of conductivity is 0.0005cal/cm-s-∘C and the inner and outer surfaces are at 0∘C and 30∘C 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 B1800 Given,
Height of refrigerator (h)=150cm
Width of refrigerator (w)=80cm
Thickness of refrigerator (d)=6cm
Coefficient of thermal conductivity (k)=0.0005cal/cm-s-∘C=0.0005×60cal/cm-min-∘C
Temperature difference between inner and outer surfaces (th−tc)=30∘C
Applying the equation of thermal conductivity, Qt=kA(th−tc)d
For t=1 minute: ⇒Q=0.0005×(60)×(150×80)×(30)6⇒Q=1800cal/min