Problem: Find the temperature of an object at the end of another half an hour if it cools from 120°F to 95°F in the first half an hour when surrounded by air whose temperature is 70°F.

Solution:
We can use Newton's Law of Cooling to solve this problem. According to this law, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.

Formula:
dT/dt = -k(T - Ts)

where dT/dt is the rate of change of temperature of the object, T is the temperature of the object, Ts is the temperature of the surroundings, and k is a constant of proportionality.

Step 1: Find the value of k.
Given, the object cools from 120°F to 95°F in half an hour.
dT/dt = (95 - 120) / (0.5) = -50°F/hour
Ts = 70°F
Using the formula, we have:
-50 = -k(120 - 70)
k = 1/10

Step 2: Find the temperature of the object after another half an hour.
Let's assume that the initial temperature of the object at t = 0 is T0.
After the first half an hour, the temperature of the object is T1 = 95°F.
Using the formula, we have:
dT/dt = -k(T - Ts)
dT/dt = -1/10 (T - 70)
Separating variables and integrating, we get:
ln(T - 70) = -1/10 t + C
where C is the constant of integration.
At t = 0, T = T0, so we have:
ln(T0 - 70) = C
Substituting T1 = 95°F and t = 0.5 hour, we get:
ln(95 - 70) = -1/10 (0.5) + ln(T0 - 70)
ln(3) = -1/20 + ln(T0 - 70)
ln(T0 - 70) = ln(3) + 1/20
T0 - 70 = e^(ln(3) + 1/20)
T0 = e^(ln(3) + 1/20) + 70
T0 = 97.9°F (approx)

After another half an hour, the temperature of the object would be:
ln(T - 70) = -1/10 (1) + ln(97.9 - 70)
ln(T - 70) = -1/10 + ln(27.9)
ln(T - 70) = 1.958
T - 70 = e^(1.958)
T = 122.1°F (approx)

Answer: The temperature of the object at the end of another half an hour would be 122.1°F.

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