The rate at which a super computer body cools is proportional to the difference between the temperature of the body and that of the surrounding air. If a body in air at 25°C will cool from 100°C to 75°C in one minute, find its temperature at the end of three minutes.
"\\int \\frac{dU}{U-25}=k\\smallint dt+ln"C
"ln(U-25)=kt+lnC"
"U-25=Ce^{kt}"
For t=0, U=100
For t=1, U=75
"100-25=Ce^{0k}"
C=75
"75-25=75e^{1k}"
"e^k=0.67"
k=-0.4
"U(3)-25=75e^{-0.4x3}"
U(3)=47.6 C
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