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Bob heats up a bottle of milk for his baby, Jill, every morning. Jill will only drink the bottle...

Question:

Bob heats up a bottle of milk for his baby, Jill, every morning. Jill will only drink the bottle if it is at 98 degrees. So Bob places the bottle in a cup of hot water (170 degrees). The bottle starts off at 35 degrees. After 2 minutes the bottle is 80 degrees. Use Newton's heating-cooling law, that the rate of change in the temperature, H, is proportional to the difference between the object and the surrounding temperature, to determine how long Bob should leave the bottle in the hot water. Assume that the water stays at 170 degrees.

Newton's Law of Heating/Cooling:

Newton's Law of Heating/Cooling states that the rate of change, {eq}\frac{dT}{dt} {/eq}, of the temperature of an object is proportional to the difference in temperature {eq}T {/eq} of the object and its surrounding, {eq}T_s {/eq}:

$$\begin{align*} &H = \frac{dT}{dt} = k (T_s - T) & \text{[Newton's Law of Heating/Cooling]} \end{align*} $$

where {eq}k {/eq} is called the object's heating/cooling constant.

Manipulating and integrating the differential equation above, when the initial temperature of the object is {eq}T_0 {/eq}, its temperature {eq}T {/eq} after some time {eq}t {/eq} is given by the equation:

$$T = T_s - (T_s - T_0) e^{(-kt)} $$

Answer and Explanation:

Given the initial temperature of the bottle {eq}T_0 = 35^\circ F {/eq}, it is placed in a cup of hot water with temperature {eq}T_s = 170^\circ F {/eq}.

Assuming the hot water stays at {eq}T_s = 170^\circ F {/eq}, the temperature of the bottle is {eq}T_2 = 80^\circ F {/eq} after {eq}t = 2 \,mins {/eq}. Determine the heating constant {eq}k {/eq} of the bottle:

{eq}\begin{align*} &T_2 = T_s - (T_s - T_0) e^{[-k(2 \,min)]} & \text{[Solve for } k ] \\ &\Rightarrow 80^\circ F = 170^\circ F - ( 170^\circ F - 35^\circ F ) e^{[-(2\,min) k]} \\ &\Rightarrow 90^\circ F = (135^\circ F) e^{[-(2\,min) k]} \\ &\Rightarrow e^{[-(2\,min) k]} = \frac{2}{3} & [ y = b^x \Leftrightarrow x = \log_b y ] \\ &\Rightarrow - (2\,min) k = \ln \left( \frac{2}{3} \right) \\ &\Rightarrow k = - \frac{1}{2 \,min} \ln \left( \frac{2}{3} \right) \\ &\Rightarrow k = 0.20273 \,min^{-1} & \text{[Heating constant of the bottle]} \end{align*} {/eq}


For the bottle to reach {eq}T_f = 98^\circ F {/eq} from {eq}T_2 = 80^\circ F {/eq}, Bob must leave the bottle in the water for an additional time {eq}t {/eq} equal to:

{eq}\begin{align*} &T_f = T_s - (T_s - T_2) e^{(-kt)} & \text{[Solve for } t ] \\ &\Rightarrow 98^\circ F = 170^\circ F - ( 170^\circ F- 80^\circ F ) e^{(-kt)} \\ &\Rightarrow 72^\circ F = ( 90^\circ F ) e^{(-kt)} \\ &\Rightarrow e^{(-kt)} = 0.8 \\ &\Rightarrow -kt = \ln 0.8 \\ &\Rightarrow t = - \frac{\ln 0.8}{k} & \text{[Substitute the value of } k ] \\ &\Rightarrow t = - \frac{\ln 0.8}{0.20273 \,min^{-1}} \\ &\Rightarrow \boxed{ t \approx 1.1 \,min } \end{align*} {/eq}


Learn more about this topic:

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What are Heating and Cooling Curves?

from College Chemistry: Help and Review

Chapter 6 / Lesson 5
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