Question

A body is heated to $110^\circ$C and placed in air at $10^\circ$C. After $1$ hour its temperature is $60^\circ$C. The additional time required for it to cool to $30^\circ$C is

• A. $\left(\dfrac{\log 2}{\log 5} + 1\right)$ hours
• B. $\left(\dfrac{\log 5}{\log 2}\right)$ hours
• C. $\left(\dfrac{\log 5}{\log 2} - 1\right)$ hours
• D. $\left(\dfrac{\log 2}{\log 5}\right)$ hours

Hint:

In Newton’s law of cooling, always subtract the surrounding temperature before applying the formula.

Answer 1

The Correct Option is C

Step 1: Using Newton’s law of cooling.
\[ T - T_s = Ce^{-kt} \] where $T_s = 10^\circ$C.
Step 2: Applying initial conditions.
At $t = 0$, $T = 110^\circ$C: \[ 100 = C \] At $t = 1$, $T = 60^\circ$C: \[ 50 = 100e^{-k} \Rightarrow e^{-k} = \frac{1}{2} \]
Step 3: Finding time to reach $30^\circ$C.
\[ 20 = 100e^{-kt} \Rightarrow e^{-kt} = \frac{1}{5} \Rightarrow t = \frac{\log 5}{\log 2} \]
Step 4: Additional time required.
\[ \text{Additional time} = \frac{\log 5}{\log 2} - 1 \]
Step 5: Conclusion.
The additional time required is $\left(\dfrac{\log 5}{\log 2} - 1\right)$ hours.