## 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(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

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1. anonymous

oh but uh uh

2. anonymous

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}$ $T=20-16e^{kt}$

3. anonymous

thanx :)