The Newton's law of cooling can be written in form
T(t)=Tout+e−kt(T0−Tout).T(t) = T_{\text{out}} + e^{-kt}(T_0-T_{\text{out}}).T(t)=Tout+e−kt(T0−Tout). Here temperature is in Kelvins.
293=278+e−k⋅30(323−278), e−30k=1545, k=0.0366.293 = 278 + e^{-k\cdot 30}(323-278), \;\; e^{-30k} = \dfrac{15}{45}, \;\; k = 0.0366.293=278+e−k⋅30(323−278),e−30k=4515,k=0.0366.
283=278+e−kt(323−278), e−kt=545, t=60 m=1 h.283 = 278 + e^{-kt}(323-278), \;\; e^{-kt} = \dfrac{5}{45}, \;\; t =60\,\mathrm{m} = 1\,\mathrm{h}.283=278+e−kt(323−278),e−kt=455,t=60m=1h.
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