Q71 / 110 · VITEEE 2023 · Mathematics

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 \(t\).

  • A
    \(\mathrm{S}-(\mathrm{S}+150) \mathrm{e}^{\mathrm{kt}}\)
  • Correct answer
    B
    \(\mathrm{S}+(150-\mathrm{S}) \mathrm{e}^{\mathrm{kt}}\)
  • C
    \(\mathrm{S}+(150+\mathrm{S}) \mathrm{e}^{\mathrm{t}}\)
  • D
    \(\mathrm{S}-(150-\mathrm{S}) \mathrm{e}^{\mathrm{kt}}\)

Solution

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 \(\mathrm{T}=\mathrm{S}+(150-\mathrm{S}) \mathrm{e}^{\mathrm{kt}}\)

Hence, the correct answer is option: 'B'.

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