Please help! Honors Algebra 2! Newton's Law of Cooling

Question

trying to check my own answers before I submit them in my math study group to study for the SAT!

This phenomenon is modeled with a differential equation, and that equation may be solved to give T(t) = TA + [T0 − TA]b^t (0 and A are bases of T), where T(t) is the varying temperature of the object at a given time, t, TA is the surrounding ambient temperature, T0 is the initial temperature of the object, and b is a constant that depends on the material the object is composed of and how fast it heats or cools. Suppose you decided to make a cup of hot tea heated to 220 F in the kitchen that is at 70 F.

1.Solve the above equation for the constant b.

2.If the cup of hot tea cooled to 170 F in 10 minutes, find the value of the constant b in the above equation. Express your answer to 5 decimal places.

3.Solve the above equation for t.

4.Suppose you like your hot tea at the tepid temperature of 100 F. How long, to the nearest minute, will you have to wait until it cools to this temperature? To go along with your hot tea, you take a frozen apple pie from the freezer and place it in the oven preheated to 350 F. Assume the freezer is at 32 F.

5.If the apple pie comes to a temperature of 100 F in 20 minutes, find the value of the constant b in the above equation. Express your answer to 5 decimal places.

6.How long will it take for the pie to reach its final temperature of 200 F?

7. The pie is taken out of the oven and set on a table in a room at 70 F. In 10 minutes, it has cooled to 180 F. However, the pie must cool to 120 F before it is ready to eat. How much longer will you have to wait?

Answer 1

1. T(t) = T_A + (T0 − TA) b^t ---> T(t) - TA = (T0 − TA) b^t ---> (T(t) - TA) / (T0 − TA) = b^t, so finally:

b = [ (T(t) - TA) / (T0 − TA) ]^(1 / t)

2. b = [ (T(t) - TA) / (T0 − TA) ]^(1 / t) = [ (170°F - 70°F) / (220°F − 70°F) ]^(1 / 10 minutes)

b = [ (100°F) / (150°F) ]^(0.1) = (2/3)^(0.1) = 0.96026

3. T(t) = TA + (T0 − TA) b^t ---> T(t) - TA = (T0 − TA) b^t ---> (T(t) - TA) / (T0 − TA) = b^t, so finally:

t = log_b[ (T(t) - TA) / (T0 − TA) ]

4. [ (T(t) - TA) / (T0 − TA) ] = log_0.96026[ (100°F - 70°F) / (220°F − 70°F) ]

t = log_0.96026[ (30°F) / (150°F) ] = log_0.96026(1/5) = 39.689 minutes = 40 minutes

5. b = [ (T(t) - TA) / (T0 − TA) ]^(1 / t) = [ (100°F - 350°F) / (32°F − 350°F) ]^(1 / 20 minutes)

b = [ (-250°F) / (−318°F) ]^(1 / 20 minutes) = (250/318)^(0.05) = 0.98804

6. t = [ (T(t) - TA) / (T0 − TA) ] = log_0.98804[ (200°F - 350°F) / (32°F − 350°F) ]

t = log_0.98804[ (-150°F) / (−318°F) ] = log_0.98804(150/318) = 62.45 minutes = 63 minutes

7. b = [ (T(t) - TA) / (T0 − TA) ]^(1 / t) = [ (180°F - 70°F) / (200°F − 70°F) ]^(1 / 10 minutes)

b = [ (110°F) / (130°F) ]^(1 / 10 minutes) = (11/13)^(0.1) = 0.98343, so we have that

t (longer to wait) = log_b[ (T(t) - T_A) / (T₀ − T_A) ] - 10 minutes

t = log_0.98343[ (120°F - 70°F) / (200°F − 70°F) ] - 10 minutes

t = log_0.98343[ (50°F) / (130°F) ] - 10 minutes = log_0.98343(5/13) - 10 minutes

t = 57.186 minutes - 10 minutes = 58 minutes - 10 minutes = 48 minutes