Question

1. susie is excited for soup season! susie makes a bowl of hot chicken noodle soup. initially the soup is 160 degrees fahrenheit. susie’s insulated bowl helps the soup cool at a rate of only 10% per minute. the soup will never get colder than room temperature at 70 degrees.

a. solve for ‘a’ and write an exponential function modeling the temperature of susie’s soup
b. using the exponential model from part a. evaluate the temperature of susie’s soup after 10 minutes.

Explanation:

Part a

Step1: Identify the formula

The general form for exponential cooling (or decay) with a horizontal asymptote (room temperature here) is \( T(t) = T_{room} + (T_0 - T_{room})a^t \), where \( T_0 \) is the initial temperature, \( T_{room} \) is room temperature, \( t \) is time in minutes, and \( a \) is the decay factor. The cooling rate is 10% per minute, so the remaining temperature factor per minute is \( 1 - 0.10 = 0.90 \), so \( a = 0.90 \).

Step2: Plug in values

Initial temperature \( T_0 = 160 \)°F, room temperature \( T_{room} = 70 \)°F. Substitute into the formula: \( T(t) = 70 + (160 - 70)(0.90)^t \). Simplify \( 160 - 70 = 90 \), so the function is \( T(t) = 70 + 90(0.90)^t \).

Step1: Use the model from part a

We have \( T(t) = 70 + 90(0.90)^t \). We need to find \( T(10) \), so substitute \( t = 10 \) into the function.

Step2: Calculate \( (0.90)^{10} \)

First, compute \( 0.90^{10} \approx 0.3486784401 \).

Step3: Compute \( 90 \times 0.3486784401 \)

\( 90 \times 0.3486784401 \approx 31.38105961 \).

Step4: Add room temperature

Add 70 to the result: \( 70 + 31.38105961 \approx 101.38 \) degrees Fahrenheit.

Answer:

\( a = 0.90 \), and the exponential function is \( T(t) = 70 + 90(0.90)^t \) (where \( T(t) \) is the temperature in degrees Fahrenheit at time \( t \) minutes).

Part b