Question

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(T-a). 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) = 20-16eKt

Answer 1

oh but uh uh

Answer 2

Separate the variables,
\[\frac{dT}{dt}=k(T-a)\]\[\frac{dT}{T-a}=kdt\]
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
\[\int\limits_{4}^{T}\frac{dT}{T-a}=k \int\limits_{0}^{t}dt\]
\[[\log (T-a)]_{4}^{T}=k[t]_{0}^{t}\]
\[\log(T-a)-\log(4-a)=k(t-0)\]\[\log(\frac{T-a}{4-a})=kt\]
\[\frac{T-a}{4-a}=e^{kt}\]
\[\frac{T-20}{4-20}=e^{kt}\]\[\frac{T-20}{-16}=e^{kt}\]\[T-20=-16e^{kt}\]
\[T=20-16e^{kt}\]

Answer 3

thanx :)