AdvancedMathematicsHigh School (11–12)Functions and CalculusNumerical

A researcher puts a thermometer, at 60 °C, into a room where a heater is slowly warming the air…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A researcher puts a thermometer, at $60\ ^\circ\text{C}$ , into a room where a heater is slowly warming the air. At $t$ hours after the start, the air temperature is $T_s = 10 + 2t$ degrees Celsius. The thermometer follows Newton's law with $k = 0.5$ per hour, that is, $\dfrac{dT}{dt} = -0.5\left(T - (10 + 2t)\right)$ . She solves it with an integrating factor and finds $T = 6 + 2t + 54e^{-0.5t}$ . She notices that the reading falls at first, reaches a lowest value, and then rises along with the room. Find the reading at $t = 4$ hours, the time at which the reading is lowest, and the lowest reading itself, all with four decimal places. Use the equation to show that at the lowest point the reading equals the air temperature at that moment.

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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