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A body cools from `60^(@)C` to `50^(@)C` in 10 minutes when kept in air at `30^(@)C` . In the next 10 minut es its temperature will beA. Below `40^(@)C`B. `40^(@)C`C. Above `40^(@)C`D. Cannot be predicted |
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Answer» Correct Answer - C `(60-50)/(10) = K [(60 + 50)/(2) - 30] "….."(1)` `(50 - theta)/(10) = K[(50+theta)/(2) - 30] "…."(2)` By `(2)//(1)` `(50-theta)/(10) = (theta - 10)/(50) = theta = 43.3^(@)C` |
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