Question

A body takes 4 min. to cool from 61$^\circ$C to 59$^\circ$C. If the temperature of the surroundings is 30$^\circ$C, the time taken by the body to cool from 51$^\circ$C to 49$^\circ$C is :

• A. 3 min.
• B. 4 min.
• C. 6 min.
• D. 8 min.

Hint:

For small temperature differences, Newton's Law of Cooling can be approximated using the average temperature of the body during the cooling interval. This makes calculations much simpler than using the integral form.

Answer 1

The Correct Option is C

To solve the given problem, we need to apply Newton's Law of Cooling. This law states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature.

Newton's Law of Cooling is mathematically expressed as:

\frac{dT}{dt} = -k(T - T_s)

where:

• T is the temperature of the body at time t.
• T_s is the surrounding temperature (30°C in this case).
• k is a positive constant.

We know from the problem that the body takes 4 minutes to cool from 61°C to 59°C. Now, let's determine the time required to cool from 51°C to 49°C.

Using the given information, let us find the constant k first:

• Initial temperature (T_1) = 61°C
• Final temperature (T_2) = 59°C
• Average temperature during the cooling process = \frac{61 + 59}{2} = 60\degree\text{C}
• Difference in temperature with surroundings = 60 - 30 = 30\degree\text{C}

According to Newton’s Law of Cooling:

T - T_s = T_1 - T_s \cdot e^{-kt}

From 61°C to 59°C in 4 minutes:

\ln\left(\frac{(61 - 30)}{(59 - 30)}\right) = kt_1

Similarly, we calculate the time from 51°C to 49°C:

• Initial temperature (T_3) = 51°C
• Final temperature (T_4) = 49°C
• Average temperature during the cooling process = \frac{51 + 49}{2} = 50\degree\text{C}
• Difference in temperature with surroundings = 50 - 30 = 20\degree\text{C}

About cooling from 51°C to 49°C:

\ln\left(\frac{(51 - 30)}{(49 - 30)}\right) = kt_2

Taking the ratio of these equations, assuming k and surrounding temperature are constant:

\frac{\ln(31/29)}{\ln(21/19)} = \frac{t_1}{t_2}

Solving the above ratio with given values:

\frac{t_1}{4} = \frac{\ln(31/29)}{\ln(21/19)}

t_2 = 6 \, \text{minutes}

Therefore, the correct option is 6 minutes.