2022 SAJC P1 Q7

2022 SAJC P1 Q7

9 marks
Prelims

The rate of temperature loss of an animal corpse can be estimated using Newton’s Law of Cooling, which states that the rate of change of temperature \(\theta {}^\circ \mathrm{C}\), \(t\) hours after death of an animal is proportional to the difference between its body temperature \(\theta {}^\circ \mathrm{C}\) and the constant surrounding temperature \({{\theta }_{0}}{}^\circ \mathrm{C}\), where \(\theta >{{\theta }_{0}}\).

  1. Write down a differential equation for this situation. Solve this differential equation and show that the general solution of the above differential equation is given by \(\theta ={{\theta }_{0}}+A{{e}^{-kt}}\), where \(A\) and \(k\) are positive constants.[3]

It is given that \({{\theta }_{0}}=24\), the initial value of \(\theta \) is \(36\) and the initial rate of temperature loss is \(2.5{}^\circ \mathrm{C}\) per hour.

  1. Calculate the exact values of \(A\) and \(k\).[3]
  2. Hence, sketch the graph of \(\theta \) against the \(t\).[2]
  3. Explain why the rate of change of temperature of an animal corpse cannot be modelled by a constant rate of decrease of \(1.5{}^\circ \mathrm{C}\) per hour.[1]

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Answer:(i) \(\dfrac{\mathrm d\theta}{\mathrm dt}=-k(\theta-\theta_0)\), \(\theta=\theta_0+A\mathrm e^{-kt}\) (ii) \(A=12\), \(k=\dfrac5{24}\) (iii) From \((0,36)\), decrease towards the asymptote \(\theta=24\) (iv) A constant decrease crosses below the surroundings after \(8\) hours and is physically unrealistic

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