A student is conducting an experiment in a laboratory to investigate how quickly liquids cool to room temperature - AQA - A-Level Maths Mechanics - Question 8 - 2019 - Paper 3

The room temperature of the laboratory can be determined from the model's limit as time approaches infinity. As $t$ gets large, the temperature $heta(t)$ approaches:

$heta = 5(4 + 0) = 20°C.$ Thus, the room temperature of the laboratory is 20°C because it is the temperature the liquid approaches after a long time.

The liquid cools to 1°C above room temperature when:

$heta = 20 + 1 = 21°C.$ Using the model:

$21 = 5(4 + e^{-kt}).$

Divide both sides by 5:

rac{21}{5} = 4 + e^{-kt} o 4.2 = 4 + e^{-kt}

Subtracting gives:

$0.2 = e^{-kt}.$ Taking the natural logarithm:

$-kt = ext{ln}(0.2).$ Substitute back for $k$:

t = -rac{ ext{ln}(0.2)}{k}. With $k ext{ approximately } 0.06806, ext{ we solve for } t,$$

$t ext{ is approximately } 58.87 ext{ minutes}.$ So, the time taken for the liquid to cool to 1°C above the room temperature is approximately 59 minutes.

The model might need to be adjusted because the cooling rates can be affected by various external conditions. For example, factors such as altitude, air flow, humidity, and the type of container used can influence the heat transfer rates significantly. Thus, if the experiment were conducted in a different environment, it may not adhere to the same cooling model established in the original experiment.