Question

recall that Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference letween the temperature of the object and the ambient temperature, i.e. T ′ =−k(T−T 0 ). (a) A cup with some liquid is placed in a fridge held at 3 C. If k is the (positive) liquid cooling constant, find the differential equation satisfied by the temperature T of the liquid. Write T for the temperature function T(t). T ′ = (b) Find the liquid temperature T as function of time knowing that the liquid initial temperature when it was placed in the fridge was 46 C. Write k for the liquid cooling constant. T(t)= (c) After 10 minutes the liquid temperature inside the fridge is 16C. Find the liquid cooling constant k. k=

Answer 1

The temperature T of the liquid satisfies the differential equation T' = -k(T - 3). The temperature as a function of time is T(t) = 3 + (46 - 3)e^(-kt) and the cooling constant is approximately 0.0579.

Step-by-step explanation:

Using Newton's law of cooling, we are given that T' = -k(T - T0) where T is the temperature of the liquid, T0 is the ambient temperature (in this case, the temperature inside the fridge), and k is the cooling constant. So the differential equation that the temperature T of the liquid satisfies is T' = -k(T - 3)

To find the temperature of the liquid as a function of time, we solve this differential equation. The solution to this is T(t) = T0 + (T(0) - T0)e^(-kt). We know that T(0) is 46°C, so we have T(t) = 3 + (46 - 3)e^(-kt).

Last, we find the liquid cooling constant k. We know that after 10 minutes (or 10/60 hours), the temperature is 16°C. Solving T(t) = 16 for k, we get k = ln((16 - 3) / (46 - 3)) / -10/60. Thus, we find the cooling constant to be approximately 0.0579.

Learn more about Newton's Law of Cooling