Question

41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.
a. How much longer would it take the soup to cool to 35°C?

Accepted Answer

Recall Newton's Law of Cooling, which states that the temperature of an object changes at a rate proportional to the difference between its temperature and the ambient temperature. Mathematically, this is expressed as: $$ T(t) = T_{ambient} + (T_0 - T_{ambient}) e^{-kt} $$ where $$T(t) is $$the temperature at time $$t$$, $$T_0 is $$the initial temperature, $$T_{ambient} is $$the room temperature, and $$k is a $$positive constant. Identify the known values from the problem: initial temperature $$T_0 = 90^\circ C$$, ambient temperature $$T_{ambient} = 20^\circ C$$, temperature after 10 minutes $$T(10) = 60^\circ C. $$Substitute these into the formula to find the constant $$k$$: $$ 60 = 20 + (90 - 20) e^{-10k} . $$Solve the equation for $$k by $$isolating the exponential term: $$ 60 - 20 = 70 e^{-10k} $$, which simplifies to $$ 40 = 70 e^{-10k} . $$Then, divide both sides by 70 and take the natural logarithm to solve for $$k$$: $$ e^{-10k} = \frac{40}{70} $$, so $$ -10k = \ln\left(\frac{40}{70}\right) $$, and finally $$ k = -\frac{1}{10} \ln\left(\frac{40}{70}\right) . $$Use the value of $$k to $$find the time $$t$$ when the soup cools to $$35^\circ C. $$Substitute $$T(t) = 35$$ into the original formula: $$ 35 = 20 + 70 e^{-kt} . $$Isolate the exponential term: $$ 15 = 70 e^{-kt} $$, then $$ e^{-kt} = \frac{15}{70} . $$Solve for $$t by $$taking the natural logarithm: $$ -kt = \ln\left(\frac{15}{70}\right) $$, so $$ t = -\frac{1}{k} \ln\left(\frac{15}{70}\right) . $$Finally, subtract the initial 10 minutes to find how much longer it takes to cool from 60°C to 35°C: $$ t_{additional} = t - 10 . $$This gives the additional cooling time needed.

41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.
a. How much longer would it take the soup to cool to 35°C?

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41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.a. How much longer would it take the soup to cool to 35°C?

Verified video answer for a similar problem:

Key Concepts

Newton’s Law of Cooling

Solving First-Order Differential Equations

Exponential Decay and Time Calculation

41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.
a. How much longer would it take the soup to cool to 35°C?