A cup of tea at 90 °C is put on a desk in a room kept at 24 °C…

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A cup of tea at $90\ ^\circ\text{C}$ is put on a desk in a room kept at $24\ ^\circ\text{C}$ . After $6$ minutes the tea is at $70\ ^\circ\text{C}$ . Assume Newton's law of cooling. Find the constant $k$ in per minute, the time in minutes at which the tea reaches $40\ ^\circ\text{C}$ , and its temperature after $15$ minutes from the start.

Need a reminder? The key ideaShow

Newton's law is $\dfrac{dT}{dt} = -k(T - T_s)$ , so $T = T_s + (T_0 - T_s)e^{-kt}$ . For a tank, $\dfrac{dQ}{dt} = r_{\text{in}}c_{\text{in}} - r_{\text{out}}\dfrac{Q}{V}$ , and the volume is $V = V_0 + (r_{\text{in}} - r_{\text{out}})t$ .

\displaystyle T - T_s = (T_0 - T_s)e^{-kt}.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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A cup of tea at 90 °C is put on a desk in a room kept at 24 °C…

The worked solution

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