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A bdoy initially at 80^(@)C ools to 64^(@)C in 5 minutes and to 52^(@)C in 10 minutes. What will be the temperature (in ""^(@)C) after 15 minutes? |
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Answer» Solution :If `T_(0)` be the temperature of the surroundings, then `-(dT)/(dt)=k(T-T_(0))` or `(80-64)/5=k((80+64)/2-T_(0))`…………I and `(80-52)/10=k((80+52)/2-T_(0))`………ii After simplifying above equations we get `T_(0)=16^(@)C` If T. be the temperature after 15 MINUTES, then `(80-T.)/15=k((80+T.)/2-16)`...........ii After simplify above equations we get `T_(0)=16^(@)C` If T. be the temperature after 15 minutes then `(80-T.)/15=k((80+T.)/2-16)`..........III From equation i and iii we get `T.=43^(@)C` |
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