Question

If \(t_1\) is the time taken for a body to cool from \(80^\circ\text{C}\) to \(75^\circ\text{C}\) and \(t_2\) is the time taken to cool from \(75^\circ\text{C}\) to \(70^\circ\text{C}\), then \[ t_1:t_2= \] \[ (\text{Temperature of surroundings}=30^\circ\text{C}) \]

• A. \(19:21\)
• B. \(17:19\)
• C. \(7:9\)
• D. \(9:11\)

Hint:

For Newton's law of cooling, \[ \boxed{ t\propto \ln\left(\frac{\theta_1}{\theta_2}\right), \qquad \theta=T-T_s. } \] Always use the excess temperature above the surroundings, not the actual temperature.

Answer 1

The Correct Option is B

Step 1: Apply Newton's law of cooling. According to Newton's law of cooling, \[ t=\frac1k \ln\left(\frac{\theta_1}{\theta_2}\right), \] where \[ \theta=T-T_s \] is the excess temperature above the surroundings. Here, \[ T_s=30^\circ\text{C}. \]

Step 2: Calculate \(t_1\). For cooling from \[ 80^\circ\text{C}\rightarrow75^\circ\text{C}, \] the excess temperatures are \[ 50^\circ\text{C} \quad\text{and}\quad 45^\circ\text{C}. \] Hence, \[ t_1=\frac1k \ln\left(\frac{50}{45}\right) = \frac1k \ln\left(\frac{10}{9}\right). \]

Step 3: Calculate \(t_2\). For cooling from \[ 75^\circ\text{C}\rightarrow70^\circ\text{C}, \] the excess temperatures are \[ 45^\circ\text{C} \quad\text{and}\quad 40^\circ\text{C}. \] Thus, \[ t_2=\frac1k \ln\left(\frac{45}{40}\right) = \frac1k \ln\left(\frac98\right). \]

Step 4: Find the ratio. Therefore, \[ \frac{t_1}{t_2} = \frac{\ln(10/9)}{\ln(9/8)}. \] Using the approximation \[ \ln(1+x)\approx x \] for small \(x\), \[ \ln\left(\frac{10}{9}\right)\approx\frac19, \] \[ \ln\left(\frac98\right)\approx\frac18. \] Hence, \[ \frac{t_1}{t_2} \approx \frac{1/9}{1/8} = \frac89 \approx \frac{17}{19}. \] Thus, \[ \boxed{t_1:t_2=17:19.} \] Therefore, the correct option is \(\boxed{(B)}\).