I need help please!

Question

1. A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 67°F room. After fifteen minutes, the internal temperature of the soup was 96°F.
2. Use Newton's Law of Cooling to write a formula T(t) that models this situation, where T is the temperature of the soup in degrees Fahrenheit and t is time in minutes.
3. A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 65°F room. After fifteen minutes, the internal temperature of the soup was 91°F.
4. To the nearest minute, how long will it take the soup to cool to 80°F?
5. A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 71°F room. After fifteen minutes, the internal temperature of the soup was 95°F.
6. To the nearest degree, what will the temperature be after 2 and a half hours?

Answer 1

1) Newton's law of cooling: T(t) = T_s + (T_o - T_s) e^-kt where T_s is temperature of surroundings and T_o the initial temperature of the cooling object. To determine the constant k, use given data: T(15 min) = 67 + (100 - 67) e^-15k = 96; 33 e^-15k = 96 - 67, e^-15k = 29/33. Take natural log of both sides, using power and identity properties: -15k * (ln e = 1) = ln(29/33) = -0.1292, k = 0.008614. So T(t) = 67 + 33 e^{-0.008614 t}.

2) Same process to find k: 91 = 65 + (100-65) e^-15k, 35 e^-15k = 26, -15k = ln(26/35), k = 0.015894. Then 80 = 65 + 35 e^-kt, 35 e^-kt = 15, -0.015894 t = ln(15/35) = -0.8473 and t ≅ 53 minutes.

3) Same process to find k: 95 = 71 + (100-71) e^-15k, 29 e^-15k = 24, -15k = ln(24/29), k = 0.012616. With t = 2.5 hours = 150 min, T(150) = 71 + 29 e^-150k = 71 + 4.37 ≅ 75 °F.