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A container is filled with a liquid that cools from 100 ""^(@)C to 70 ""^(@)C. The times that it must have taken to cool down to 80 ""^(@)C from its initial temperature approximately is |
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Answer» 1.7 min In first case, `{:(T_1=100^@C),(T_2=70^@C):}""|""{:(T_0=30^@C),("time,"t_1=5min=300s):}` According to Newton. s law of cooling, `mc((T_(1)-T_(2))/(t))=K ((T_(1)+T_(2))/(2)-T_(0))` `mc[(100-70)/(300)] = k [(100+70)/(2)-30]` `mc[(30)/(300)] =k[(170)/(2)-30]` `mc[(1)/(10)] = k [85-30]` `rArr mc ((1)/(10)) = k(55)` For, second case, `{:(T_1=100^@C),(T_2=80^@C):}""|""{:(T_0=30^@C),("time,"t_2=t.min=t.s):}` According to Newton. s law of cooling , `mc[(100-80)/(t.)]=k [(100+80)/(2)-30]` `mc[(20)/(t.)]=k[90-30]` `rArr mc[(20)/(t.)]=k[60]` After, solving Eqs: (i) and (ii), we get t. = 3. 05 min (NEAREST answer isb) |
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