When a cold drink is taken from a refrigerator, its temperature is 5 ^{\circ}C. After 25 minutes...

Question:

When a cold drink is taken from a refrigerator, its temperature is 5 {eq}^{\circ} {/eq}C. After 25 minutes in a 20 {eq}^{\circ} {/eq}C room its temperature has increased to 10 {eq}^{\circ} {/eq}C. (Round your answer to two decimal places)

When will its temperature be 14 {eq}^{\circ} {/eq}C?

Newton's Law of Heating/Cooling:

The Newton's Law of Heating/Cooling states that the rate of change of the temperature of an object is directly proportional to the difference of its own temperature {eq}T {/eq} and the temperature of its surroundings {eq}T_s {/eq}.

$$\frac{dT}{dt} = - k (T - T_s) $$

where {eq}k {/eq} is the heating/cooling constant of the object. Manipulating and integrating the equation above, the temperature {eq}T {/eq} of an object after some time {eq}t {/eq} when its initial temperature is {eq}T_0 {/eq} is given by the equation:

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

Answer and Explanation:

Taken from a refrigerator, the initial temperature of a cold drink is given to be {eq}T_0 = 5^\circ C {/eq}.

After being exposed to a room with temperature {eq}T_s = 20^\circ C {/eq} for {eq}t = 25 \,mins {/eq}, the temperature of the drink increased to {eq}T(25) = 10^\circ {/eq}.

First, determine the heating/cooling constant of the cold drink:

{eq}\begin{align*} &T(t) = T_s + ( T_0 - T_s ) e^{-kt} & \text{[Solve for } k ] \\ &\Rightarrow T(t) - T_s = ( T_0 - T_s ) e^{-kt} \\ &\Rightarrow e^{-kt} = \frac{ T(t) - T_s }{ T_0 - T_s } & \left[ y = b^x \Leftrightarrow x = \log_b y \right] \\ &\Rightarrow -kt = \ln \left( \frac{ T(t) - T_s }{ T_0 - T_s } \right) & \text{[Equation 1]} \\ \\ &\Rightarrow k = - \frac{1}{t} \ln \left( \frac{ T(t) - T_s }{ T_0 - T_s } \right) & \text{[Substitute the given values when } t = 25 \,mins ] \\ &\Rightarrow k = - \frac{1}{25 \,mins} \ln \left( \frac{ 10^\circ C - 20^\circ C }{ 5^\circ C - 20^\circ C } \right) \\ &\Rightarrow k = - \frac{1}{25} \ln \left( \frac{ - 10^\circ C }{ - 15^\circ C } \right) \,mins^{-1} \\ &\Rightarrow k = - \frac{1}{25} \ln \left( \frac{2}{3} \right) \,mins^{-1} & \text{[Heating/cooling constant of the cold drink]} \end{align*} {/eq}


Finally, using Equation 1, determine the time {eq}t {/eq} needed for the cold drink to be exposed to the room temperature to have its temperature raised to {eq}T = 14^\circ C {/eq}:

{eq}\begin{align*} &-kt = \ln \left( \frac{ T(t) - T_s }{ T_0 - T_s } \right) & \text{[Solve for } t ] \\ &\Rightarrow t = - \frac{1}{k} \ln \left( \frac{ T(t) - T_s }{ T_0 - T_s } \right) & \text{[Substitute all known values]} \\ &\Rightarrow t = - \left( - \frac{1}{25} \ln \left( \frac{2}{3} \right) \,mins^{-1} \right)^{-1} \ln \left( \frac{ 14^\circ C - 20^\circ C }{ 5^\circ C - 20^\circ C } \right) \\ &\Rightarrow t = \frac{ 25 \ln \left( \frac{-6}{-15} \right) }{ \ln \left( \frac{2}{3} \right) } \,mins \\ &\Rightarrow t = \frac{ 25 \ln \left( \frac{2}{5} \right) }{ \ln \left( \frac{2}{3} \right) } \,mins \\ &\Rightarrow \boxed{ t = 56.50 \,mins } & \boxed{ \text{Time needed for the } 5^\circ C \text{-cold drink to reach the temperature of } 14^\circ C \text{ in a } 20^\circ C \text{ room} } \end{align*} {/eq}


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