Question

A heat exchanger during operation in a bioprocess has a steady temperature of 90 \(^{\circ}\)C. After completion of its operation, it was shut down and it was observed that the rate of decrease of temperature at any time was directly proportional to the difference \(T(t) - 30\) \(^{\circ}\)C, where \(T(t)\) denotes temperature at time \(t\). It was observed that it took 30 min for the temperature to drop to 70 \(^{\circ}\)C. The temperature after 51.5 min will be ________ \(^{\circ}\)C. (rounded off to the nearest integer)

Hint:

Use Newton's law of cooling, \(T(t) = 30 + 60e^{-kt}\); find \(k\) from the 30 min data point, then evaluate at \(t=51.5\) min.

Answer 1

Correct Answer: 60

Step 1: Set up the governing differential equation.
Newton's law of cooling says the rate of temperature drop is proportional to the difference between the object's temperature and the surrounding temperature (here 30 \(^{\circ}\)C):
\[ \frac{dT}{dt} = -k\left(T(t) - 30\right) \]
where \(k > 0\) is the cooling rate constant.

Step 2: Solve the differential equation.
Separating variables and integrating from \(t=0\) (where \(T=90\)) gives the standard exponential decay solution:
\[ T(t) - 30 = (90 - 30)e^{-kt} \]
\[ T(t) = 30 + 60e^{-kt} \]

Step 3: Use the given data point to find \(k\).
At \(t = 30\) min, \(T = 70\) \(^{\circ}\)C:
\[ 70 = 30 + 60e^{-30k} \]
\[ 40 = 60e^{-30k} \]
\[ e^{-30k} = \frac{2}{3} \]
\[ -30k = \ln\left(\frac{2}{3}\right) \]
\[ k = \frac{\ln(3/2)}{30} = \frac{0.405465}{30} = 0.013516\ \text{min}^{-1} \]

Step 4: Find the temperature at \(t = 51.5\) min.
\[ T(51.5) = 30 + 60\,e^{-0.013516 \times 51.5} = 30 + 60\,e^{-0.696048} \]
Since \(e^{-0.693147} = 0.5\) exactly (as \(\ln 2 = 0.693147\)),
\[ e^{-0.696048} \approx 0.5 \times e^{-0.002901} \approx 0.5 \times 0.997102 \approx 0.498551 \]
\[ T(51.5) \approx 30 + 60(0.498551) \approx 30 + 29.913 \approx 59.91 \]

Final Answer:
Rounded off to the nearest integer,
\[ \boxed{T(51.5\ \text{min}) \approx 60\ ^{\circ}\text{C}} \]