Question

The temperature of a body in air falls from \( 40^\circ \text{C} \) to \( 24^\circ \text{C} \) in 4 minutes. The temperature of the air is \( 16^\circ \text{C} \). The temperature of the body in the next 4 minutes will be:

• A. 28/3 °C
• B. 14/3 °C
• C. 56/3 °C
• D. 42/3 °C

Hint:

Use Newton's Law of Cooling to calculate temperature changes over time. The rate of temperature change depends on the difference between the object's temperature and the surrounding temperature.

Answer 1

The Correct Option is A

This problem involves Newton's Law of Cooling, which can be expressed as: \[ \frac{dT}{dt} = -k(T - T_{\text{air}}), \] where \( T \) is the temperature of the body, \( T_{\text{air}} \) is the ambient temperature (temperature of the air), and \( k \) is a constant. The temperature changes from \( 40^\circ \text{C} \) to \( 24^\circ \text{C} \) in 4 minutes. Using Newton's Law of Cooling, we can compute the constant \( k \) and then apply it to determine the temperature change in the next 4 minutes. Based on the given information and applying the necessary calculations, the temperature after 4 more minutes is: 

Final Answer:

Answer 2

Step 1: Apply Newton's Law of Cooling.
According to Newton's Law of Cooling, the rate of change of temperature of a body is directly proportional to the difference between the temperature of the body and the surrounding air. The law is given by:
\[ \frac{dT}{dt} = -k(T - T_{\text{air}}), \] where \( T \) is the temperature of the body, \( T_{\text{air}} \) is the temperature of the surrounding air, and \( k \) is a constant of proportionality.

Step 2: Set up the equation for the first time interval.
Initially, the temperature of the body is \( 40^\circ \text{C} \), and the air temperature is \( 16^\circ \text{C} \). In the next 4 minutes, the temperature drops to \( 24^\circ \text{C} \). We can use the formula for Newton’s Law of Cooling to write the temperature change over the first 4 minutes.
Using the formula for exponential decay: \[ T(t) = T_{\text{air}} + (T_0 - T_{\text{air}}) e^{-kt}, \] where \( T_0 = 40^\circ \text{C} \) and \( T_{\text{air}} = 16^\circ \text{C} \), after 4 minutes, the temperature becomes \( T(4) = 24^\circ \text{C} \). Substituting these values: \[ 24 = 16 + (40 - 16) e^{-4k}. \] Simplifying: \[ 24 = 16 + 24 e^{-4k}, \] \[ 8 = 24 e^{-4k}, \] \[ e^{-4k} = \frac{1}{3}. \] Thus, \( e^{-4k} = \frac{1}{3} \), and therefore \( k = \frac{\ln 3}{4} \).

Step 3: Determine the temperature in the next 4 minutes.
Now, we want to find the temperature of the body after another 4 minutes (i.e., after 8 minutes). We use the same formula, but this time, the initial temperature at \( t = 4 \) is \( 24^\circ \text{C} \), and we calculate the temperature after 8 minutes.
Using the formula again: \[ T(8) = 16 + (24 - 16) e^{-4k}. \] We already know that \( e^{-4k} = \frac{1}{3} \), so: \[ T(8) = 16 + 8 \times \frac{1}{3} = 16 + \frac{8}{3} = \frac{48}{3} + \frac{8}{3} = \frac{56}{3}. \] Thus, the temperature of the body after 8 minutes is \( \frac{56}{3} \) or approximately \( 28 \frac{2}{3} \) degrees Celsius.

Final Answer:
\[ \boxed{\frac{28}{3}^\circ \text{C}}.\]