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The specific heat of a substance varies as (3t^(2)+t)xx10^(-3)cal//g""^(@)C. What is the amount of heat required to raise the temperature of 1 kg of substance from 10^(@)C" to "20^(@)C ? |
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Answer» Solution :For small change in temperature dt, heat REQUIRED, `dQ=MSDT" "thereforeQ=int_(t_(1))^(t_(2))msdt` Given , m = 1000 g, s = `(3t^(2)+t)` `thereforeQ=int_(10)^(20)1000(3t^(2)+t)xx10^(-3)dt` = `|t^(3)+t^(2)/2|_(10)^(20)=(20^(3)+20^(2)/2)-(10^(3)+10^(2)/2)=8200-1050` = 7150 CAL |
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