Question

The temperature of the body drops from 60°C to 40°C in 7 min. The surrounding temperature is 10°C. The temperature of the body drops from 40°C to T°C in 7 min. Find the value of T

• A. 16°C
• B. 20°C
• C. 28°C
• D. 36°C

Answer 1

The Correct Option is C

To solve this problem, we will use Newton's Law of Cooling, which is expressed by the formula:

\(T(t) = T_s + (T_0 - T_s) e^{-kt}\)

Where:

• \(T(t)\) is the temperature of the body at time \(t\).
• \(T_s\) is the surrounding temperature.
• \(T_0\) is the initial temperature of the body.
• \(k\) is a constant.
• \(t\) is the time.

Given:

• The initial temperature \(T_0 = 60^\circ C\).
• The final temperature after 7 minutes is \(40^\circ C\).
• The surrounding temperature is \(T_s = 10^\circ C\).

Using the formula for the first part of cooling (from \(60^\circ C\) to \(40^\circ C\)):

\(40 = 10 + (60 - 10) e^{-7k}\)

Simplifying, we get:

\(30 = 50 e^{-7k}\)

\(\frac{30}{50} = e^{-7k}\)

\(e^{-7k} = 0.6\) (Equation 1)

For the second phase of cooling (from \(40^\circ C\) to \(T^\circ C\) in the next 7 minutes):

\(T = 10 + (40 - 10) e^{-7k}\)

Since we know \(e^{-7k} = 0.6\) from Equation 1, substitute this into the new equation:

\(T = 10 + 30 \times 0.6\)

\(T = 10 + 18 = 28^\circ C\)

Therefore, the temperature \(T\) after the second 7 minutes is 28°C.