2025 DHS P2 Q5

2025 DHS P2 Q5

10 marks

A heated metal block at \(160^\circ C\) is left to cool in a laboratory with an ambient temperature of \(30^\circ C\). The temperature, \(T\) in \(^\circ C\), of the metal block \(t\) minutes after it is left in the laboratory is model as a function \(\mathrm{T}(t)\). The rate of cooling is proportional to the difference between the object's temperature and the ambient temperature, with a positive constant of proportionality \(k\).

  1. Write down a differential equation for the situation and find the expression of \(\mathrm{T}(t)\) in terms of \(k\).[3]

An experiment to study the cooling of the heated metal block starts at 10:00 am. However, a protective thermal casing is used to delay the cooling of the metal block for \(5\) minutes, resulting in the temperature of the metal block remaining constant at \(160^\circ C\) during this time. At 10:05 am, the casing is removed and cooling begins immediately. At 10:15 am, the temperature of the metal block is measured to be \(100^\circ C\).

  1. In order to use the solution obtained in part (a) to model the cooling process now, the time \(t\) needs to be replaced with \(t+a\), where \(a\) is a constant. State the value of \(a\).[1]
  2. Due to the protective thermal casing, the temperature of the metal block \(T_p(t)\) \(^\circ C\), \(t\) minutes after 10:00 am can be modelled by the following piecewise function \[T_p(t) = \begin{cases} 160 & \text{for}\hspace{0.5em} 0 \le t < 5, \\ T(t+a) & \text{for}\hspace{0.5em} t \ge 5. \end{cases}\]
    Using your answers from parts (a) and (b), show that \(T(t+a) = 30 + 177\mathrm{e}^{-0.0619t}\).[3]
  3. Find the time it takes for the metal block to cool to \(60^\circ C\).[1]
  4. Sketch the graph of \(T_p(t)\) against \(t\) for \(t \ge 0\).[2]

Solution:

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Answer:(a) \(T^{\prime}=-k(T-30)\), \(T=30+130e^{-kt}\). (b) \(a=-5\). (c) \(T(t-5)\approx30+177e^{-0.0619t}\). (d) \(28.7\) minutes after 10:00 am. (e) Plateau160°C to5 minutes, then decreasing towards30°C; sketch shown.

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