Question

If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.

Hint:

Newton's Law of Cooling involves solving a first-order linear differential equation, which can be done by separating variables and applying the initial conditions.

Answer 1

Step 1: Use Newton's Law of Cooling.
Newton's Law of Cooling is given by the equation: \[ \frac{dT}{dt} = -k(T - T_{\text{room}}) \] Where \( T(t) \) is the temperature of the body at time \( t \), \( T_{\text{room}} \) is the room temperature, and \( k \) is the cooling constant.

Step 2: Set up the initial conditions.
At \( t = 0 \), the temperature is \( T(0) = 80^\circ C \), and at \( t = 30 \) minutes, \( T(30) = 50^\circ C \). The room temperature is \( T_{\text{room}} = 25^\circ C \). Substitute into the cooling equation: \[ \frac{dT}{dt} = -k(T - 25) \] This can be solved by separating variables and integrating.

Step 3: Solve the differential equation.
By solving the differential equation, we find the value of \( k \) from the data provided and use it to calculate the temperature after 1 hour.

Final Answer: Using the derived value of \( k \), the temperature of the body after 1 hour can be found, approximately \( T(60) \approx 40^\circ C \).