Question

A body cools from $60^\circ\text{C}$ to $40^\circ\text{C}$ in 6 minutes. After next 6 minutes its temperature will be (Temperature of the surroundings is $10^\circ\text{C}$ )}

• A. $24^\circ\text{C}$
• B. $28^\circ\text{C}$
• C. $18^\circ\text{C}$
• D. $32^\circ\text{C}$

Hint:

Cooling rates slow down as the body gets closer to the surrounding temperature. The drop in the second interval will always be less than the first.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its own average temperature and the ambient temperature of its surroundings.
Step 2: Key Formula or Approach:
The approximate form of Newton's law of cooling is:
\[ \frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) \]
where $T_1$ is the initial temperature, $T_2$ is the final temperature, $t$ is the time taken, $T_s$ is the surrounding temperature, and $K$ is a cooling constant.
Step 3: Detailed Explanation:
Given $T_s = 10^\circ\text{C}$.
Case 1: First 6 minutes
$T_1 = 60^\circ\text{C}$, $T_2 = 40^\circ\text{C}$, $t = 6\text{ mins}$.
Substitute these into the formula to find $K$:
\[ \frac{60 - 40}{6} = K \left( \frac{60 + 40}{2} - 10 \right) \]
\[ \frac{20}{6} = K (50 - 10) \]
\[ \frac{10}{3} = 40K \implies K = \frac{10}{3 \times 40} = \frac{1}{12}\text{ min}^{-1} \]
Case 2: Next 6 minutes
Now the initial temperature is $T_1 = 40^\circ\text{C}$. Let the new final temperature be $T$. $t = 6\text{ mins}$.
Substitute the known values and the calculated $K$:
\[ \frac{40 - T}{6} = \frac{1}{12} \left( \frac{40 + T}{2} - 10 \right) \]
Simplify the right side:
\[ \frac{40 - T}{6} = \frac{1}{12} \left( \frac{40 + T - 20}{2} \right) \]
\[ \frac{40 - T}{6} = \frac{1}{12} \left( \frac{20 + T}{2} \right) = \frac{20 + T}{24} \]
Cross-multiply to solve for $T$:
\[ 24(40 - T) = 6(20 + T) \]
Divide both sides by 6:
\[ 4(40 - T) = 20 + T \]
\[ 160 - 4T = 20 + T \]
\[ 160 - 20 = T + 4T \]
\[ 140 = 5T \]
\[ T = \frac{140}{5} = 28^\circ\text{C} \]
Step 4: Final Answer:
The temperature after the next 6 minutes will be $28^\circ\text{C}$.