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The temperature \(T\) of a cooling object drops at a rate proportional to the difference \(\mathrm{T}-\mathrm{S}\), where \(\mathrm{S}\) is constant temperature of surrounding medium. If initially \(\mathrm{T}=150^{\circ} \mathrm{C}\), find the temperature of the cooling object at any time \(t\).
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The correct answer is:
\(\mathrm{S}+(150-\mathrm{S}) \mathrm{e}^{\mathrm{kt}}\)
Let \(\mathrm{T}\) be the temperature of the cooling object at any time \(t\).
\(\frac{\mathrm{dT}}{\mathrm{dt}} \propto(\mathrm{T}-\mathrm{S}) \Rightarrow \frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{k}(\mathrm{T}-\mathrm{S}) \Rightarrow \mathrm{T}-\mathrm{S}=\mathrm{ce}^{\mathrm{kt}},\)
where \(\mathrm{k}\) is negative.
\(\Rightarrow \mathrm{T}=\mathrm{S}+\mathrm{ce}^{\mathrm{kt}}\)
When \(\mathrm{t}=0, \mathrm{~T}=150 \Rightarrow 150=\mathrm{S}+\mathrm{c}\)
\(\Rightarrow \mathrm{c}=150-\mathrm{S}\)
\(\therefore\) The temperature of the cooling object at any time is
\(\mathrm{T}=\mathrm{S}+(150-\mathrm{S}) \mathrm{e}^{\mathrm{kt}}\)
\(\frac{\mathrm{dT}}{\mathrm{dt}} \propto(\mathrm{T}-\mathrm{S}) \Rightarrow \frac{\mathrm{dT}}{\mathrm{dt}}=\mathrm{k}(\mathrm{T}-\mathrm{S}) \Rightarrow \mathrm{T}-\mathrm{S}=\mathrm{ce}^{\mathrm{kt}},\)
where \(\mathrm{k}\) is negative.
\(\Rightarrow \mathrm{T}=\mathrm{S}+\mathrm{ce}^{\mathrm{kt}}\)
When \(\mathrm{t}=0, \mathrm{~T}=150 \Rightarrow 150=\mathrm{S}+\mathrm{c}\)
\(\Rightarrow \mathrm{c}=150-\mathrm{S}\)
\(\therefore\) The temperature of the cooling object at any time is
\(\mathrm{T}=\mathrm{S}+(150-\mathrm{S}) \mathrm{e}^{\mathrm{kt}}\)
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