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A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^(@)C`. Initial temeprature of cooling is valid. The body cools to `38^@C` in 10 min In further 10 min it will cools from `38^@C` to _____A. `36^@C`B. `36.4^@C`C. `37^@C`D. `37.5^@C` |
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Answer» Correct Answer - B We have `theta-theta_s=(theta_0-theta_s)e^(-kt)` Where `theta_0=`initial temperature of body `=40^@C` `theta=`temperature of body after time t. Since body cools from 40 to 38 in 10 min, we have `38-30=(40-30)e^(-10k)` .(i) Let after 10 min, the body temperature be `theta` `theta-30^@=(38-30)e^(-10k)` .(ii) `(Eq(i))/(Eq(ii))`gives `(8)/(theta-30)=(10)/(8)`,`theta-30=6.4` `theta=36.4^@C` |
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