A cup of hot tea is sitting in a 20 ^{\circ}C room. After cooling for 10 minutes, it is 75...

Question:

A cup of hot tea is sitting in a 20 {eq}^{\circ}{/eq}C room. After cooling for 10 minutes, it is 75 {eq}^{\circ}{/eq}C. And, after 30 minutes (total) sitting in the room, it is 47 {eq}^{\circ}{/eq}C. What was the temperature of the tea when it was first put in the room?

Newton's Law of Cooling:

Given a substance with initial temperature {eq}T_0 {/eq} exposed to the surroundings with temperature {eq}T_s < T_0 {/eq}.

After some time {eq}t {/eq}, the temperature of the substance is given by the equation:

$$\begin{align*} &T = T_s + (T_0 - T_s) e^{-kt} & \text{[Newton's Law of Cooling]} \end{align*} $$

where {eq}k {/eq} is the cooling constant of the substance. Note that the temperatures must either be all in Celsius or all in Kelvins.

Answer and Explanation:

Given a cup of hot tea sitting in a room temperature of {eq}T_s = 20^\circ C {/eq}.

After cooling for {eq}t_1 = 10 \,min {/eq}, the temperature of...

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