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Earth receives `1400 W//m^2` of solar power. If all the solar energy falling on a lens of area `0.2m^2` is focused on to a block of ice of mass 280 grams, the time taken to melt the ice will be….. Minutes. (`Latent heat of fusion of ice`=`3.3xx10^5J//kg`.) |
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Answer» Solar power received by earth `=1400W//m^(2)` Solar power received by `0.2m^(2)` area `=(1400W//m^(2))(0.2m^(2))=280W` Mass of ice `=280g=0.280Kg` Here rerquired to melt ice `=(0.280)(3.3xx10^(5))` `=9.24xx10^(4)J` If `t` is the time taken for the ice to melt, we will have `(280)t=9.24xx10^(4)J[:. P=(E)/(t)]` `t=(9.24xx10^(4))/(280)s=330s=5.5min` |
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