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anonymous
 one year ago
we take milk carton out of the refrigerator and put it on the table in room temp. we assume that the temperature of the carton, T, as a function of time t. following Newtons cooling law dT/dt = k(Ta). where a is the environmental temperature, and k is Konstan.
we assume that the temperature of the environment is a = 20 and T (0) = 4.
show that T (t) = 2016eKt
anonymous
 one year ago
we take milk carton out of the refrigerator and put it on the table in room temp. we assume that the temperature of the carton, T, as a function of time t. following Newtons cooling law dT/dt = k(Ta). where a is the environmental temperature, and k is Konstan. we assume that the temperature of the environment is a = 20 and T (0) = 4. show that T (t) = 2016eKt

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Separate the variables, \[\frac{dT}{dt}=k(Ta)\]\[\frac{dT}{Ta}=kdt\] Now you are given the limits \[t=0 \space \& \space T=4 \space \space \space ; \space \space \space t=t \space \& \space T=T\] So integrate within the limits \[\int\limits_{4}^{T}\frac{dT}{Ta}=k \int\limits_{0}^{t}dt\] \[[\log (Ta)]_{4}^{T}=k[t]_{0}^{t}\] \[\log(Ta)\log(4a)=k(t0)\]\[\log(\frac{Ta}{4a})=kt\] \[\frac{Ta}{4a}=e^{kt}\] \[\frac{T20}{420}=e^{kt}\]\[\frac{T20}{16}=e^{kt}\]\[T20=16e^{kt}\] \[T=2016e^{kt}\]
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