A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 18° F. After one- half minute the thermometer reads 59° F. What is the reading of the thermometer at t = 1 min? Assume that the rate of change of temperature in time is proportional to the difference between the thermometer's temperature and the air temperature.
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 18° F. After one- half minute the thermometer reads 59° F. What is the reading of the thermometer at t = 1 min? Assume that the rate of change of temperature in time is proportional to the difference between the thermometer's temperature and the air temperature.
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![**Example Problem: Newton's Law of Cooling**
**Problem Statement:**
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 18° F. After one-half minute the thermometer reads 59° F. What is the reading of the thermometer at \( t = 1 \) min? Assume that the rate of change of temperature in time is proportional to the difference between the thermometer's temperature and the air temperature.
**Instructions:**
Type your answer in °F in the space provided below. Round your answer to two decimal places.
**Detailed Explanation:**
Apply the Newton's Law of Cooling formula:
\[ T(t) = T_s + (T_0 - T_s) e^{-kt} \]
Where:
- \( T(t) \) is the temperature of the object at time \( t \).
- \( T_s \) is the surrounding temperature.
- \( T_0 \) is the initial temperature of the object.
- \( k \) is the cooling constant.
- \( t \) is the time in minutes.
Given:
- Initial Room Temperature, \( T_0 = 70° F \)
- Surrounding Air Temperature, \( T_s = 18° F \)
- At \( t = 0.5 \) minutes, the thermometer reads \( T(0.5) = 59° F \).
1. First, use the given information to find the cooling constant \( k \).
At \( t = 0.5 \):
\[ 59 = 18 + (70 - 18)e^{-0.5k} \]
2. Solve for \( k \):
\[ 59 = 18 + 52e^{-0.5k} \]
\[ 41 = 52e^{-0.5k} \]
\[ \frac{41}{52} = e^{-0.5k} \]
\[ \ln\left(\frac{41}{52}\right) = -0.5k \]
\[ k = -2 \ln\left(\frac{41}{52}\right) \]
3. Substitute \( k \) back into the Newton's Law of Cooling formula to find the temperature at \( t = 1 \) minute:
\[ T(1) = 18 + 52e^{-k} \]
**Answer:**
Type your answer in °F in the space](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2e521a1e-f4ee-4c02-af71-8dfc2cc184d3%2Fa997620f-742b-4f59-9565-d84715097186%2Fv88apz_processed.png&w=3840&q=75)
Transcribed Image Text:**Example Problem: Newton's Law of Cooling**
**Problem Statement:**
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 18° F. After one-half minute the thermometer reads 59° F. What is the reading of the thermometer at \( t = 1 \) min? Assume that the rate of change of temperature in time is proportional to the difference between the thermometer's temperature and the air temperature.
**Instructions:**
Type your answer in °F in the space provided below. Round your answer to two decimal places.
**Detailed Explanation:**
Apply the Newton's Law of Cooling formula:
\[ T(t) = T_s + (T_0 - T_s) e^{-kt} \]
Where:
- \( T(t) \) is the temperature of the object at time \( t \).
- \( T_s \) is the surrounding temperature.
- \( T_0 \) is the initial temperature of the object.
- \( k \) is the cooling constant.
- \( t \) is the time in minutes.
Given:
- Initial Room Temperature, \( T_0 = 70° F \)
- Surrounding Air Temperature, \( T_s = 18° F \)
- At \( t = 0.5 \) minutes, the thermometer reads \( T(0.5) = 59° F \).
1. First, use the given information to find the cooling constant \( k \).
At \( t = 0.5 \):
\[ 59 = 18 + (70 - 18)e^{-0.5k} \]
2. Solve for \( k \):
\[ 59 = 18 + 52e^{-0.5k} \]
\[ 41 = 52e^{-0.5k} \]
\[ \frac{41}{52} = e^{-0.5k} \]
\[ \ln\left(\frac{41}{52}\right) = -0.5k \]
\[ k = -2 \ln\left(\frac{41}{52}\right) \]
3. Substitute \( k \) back into the Newton's Law of Cooling formula to find the temperature at \( t = 1 \) minute:
\[ T(1) = 18 + 52e^{-k} \]
**Answer:**
Type your answer in °F in the space
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