Question

the liquid base of an ice cream has an initial temperature of begin mathsize 24px style bold 93 to the power of bold omicron bold italic c end style before it is placed in a freezer with a constant temperature of begin mathsize 28px style bold minus bold 19 to the power of bold omicron bold italic c bold. end style after 1 hour, the temperature of the liquid base has decreased to begin mathsize 28px style bold 57 to the power of bold omicron bold italic c end style. use the newton's law of cooling to formulate and solve this initial value problem to find the temperature of the liquid base 2 hours after it was placed in the freezer in degrees c. round your answer to two decimal places.

Answer 1

According to Newton's Law of Cooling, the temperature of a substance decreases over time when it is in contact with a colder surrounding environment. To find the temperature of the liquid base 2 hours after being placed in the freezer, we can use the formula for Newton's Law of Cooling and the given initial temperature and surrounding temperature.

Step-by-step explanation:

According to Newton's Law of Cooling, the rate of change of temperature of a substance is directly proportional to the temperature difference between the substance and its surroundings. The formula for Newton's Law of Cooling is:

T(t) = T0 + (T1 - T0)e-kt

Where T(t) is the temperature at time t, T0 is the initial temperature, T1 is the temperature of the surroundings, and k is a constant. To find the temperature of the liquid base 2 hours after it was placed in the freezer, we can use the given information:

Initial temperature (T0) = 93°C

Temperature of the surroundings (T1) = -19°C

Time (t) = 2 hours

Plugging these values into the formula, we get:

T(2) = 93 + (-19 - 93)e-k(2)

To find the constant k, we can use the fact that the temperature of the liquid base decreased to 57°C after 1 hour:

57 = 93 + (-19 - 93)e-k(1)

Solving this equation will give us the value of k. Once we have the value of k, we can substitute it back into the formula to find the temperature at t=2 hours.