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JEE Main & Advanced Physics Transmission of Heat Question Bank Critical Thinking

  • question_answer
    An ice box used for keeping eatable cold has a total wall area of \[1\ metr{{e}^{2}}\] 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]

    A)            \[776\ gms\]                        

    B)            \[7760\ gms\]

    C)            \[11520\ gms\]                    

    D)            \[1552\ gms\]

    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 \[Q=mL\] Þ \[m=\frac{Q}{L}\]\[=\frac{518400}{334\times {{10}^{3}}}\]             = 1.552 kg = 1552g.


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