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A cup of coffee cools from 88 °C to 78 °C in 3.5 minutes in a room at 22 °C…

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A cup of coffee cools from $88\ ^\circ\text{C}$ to $78\ ^\circ\text{C}$ in $3.5$ minutes in a room at $22\ ^\circ\text{C}$ . Find the cooling constant $k$ , and when the coffee reaches $50\ ^\circ\text{C}$ .

Need a reminder? The key ideaShow

Newton's law of cooling: $\dfrac{dT}{dt} = -k(T - T_0)$ , so $T = T_0 + (T_s - T_0)e^{-kt}$ . Over a short interval, the rate is in proportion to the average excess over the surroundings.

The worked solution

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