Math Problem Statement
type this nicely so when i copy and paste it will look correctly and have exponents on the top. : Newton’s Law of Cooling
The formula:
T ( t )
T r + ( T i − T r ) e k t T(t)=T r +(T i −T r )e kt
where:
T ( t ) T(t): Temperature at time t t T r T r : Room temperature T i T i : Initial temperature k k: Cooling constant t t: Time Given:
Room temperature ( T r T r ) = 7 0 ∘ 70 ∘
Initial temperature ( T i T i ) = 18 0 ∘ 180 ∘
At t
5 t=5 minutes, T ( 5 )
14 0 ∘ T(5)=140 ∘
Step 1: Determine the cooling constant k k:
From the formula:
T ( t )
T r + ( T i − T r ) e k t T(t)=T r +(T i −T r )e kt
Substitute the known values:
140
70 + ( 180 − 70 ) e 5 k 140=70+(180−70)e 5k
Simplify:
140
70 + 110 e 5 k 140=70+110e 5k
70
110 e 5 k 70=110e 5k
70 110
e 5 k 110 70 =e 5k
7 11
e 5 k 11 7 =e 5k
Take the natural logarithm ( ln ln) on both sides:
ln ( 7 11 )
5 k ln( 11 7 )=5k Solve for k k:
k
ln ( 7 11 ) 5 k= 5 ln( 11 7 )
Using a calculator:
k ≈ ln ( 0.6364 ) 5 ≈ − 0.452 5 ≈ − 0.0904 k≈ 5 ln(0.6364) ≈ 5 −0.452 ≈−0.0904 Thus:
k ≈ − 0.0904 k≈−0.0904 Step 2a: Find the temperature after 10 minutes ( t
10 t=10):
Using the formula:
T ( t )
T r + ( T i − T r ) e k t T(t)=T r +(T i −T r )e kt
Substitute the known values:
T ( 10 )
70 + ( 180 − 70 ) e − 0.0904 ⋅ 10 T(10)=70+(180−70)e −0.0904⋅10
T ( 10 )
70 + 110 e − 0.904 T(10)=70+110e −0.904
Using a calculator:
e − 0.904 ≈ 0.405 e −0.904 ≈0.405 T ( 10 )
70 + 110 ( 0.405 ) T(10)=70+110(0.405) T ( 10 ) ≈ 70 + 44.55 ≈ 114.5 5 ∘ T(10)≈70+44.55≈114.55 ∘
Thus:
T ( 10 ) ≈ 114. 6 ∘ T(10)≈114.6 ∘
Step 2b: Find the time ( t t) for the temperature to cool to 9 5 ∘ 95 ∘ :
Using the formula:
T ( t )
T r + ( T i − T r ) e k t T(t)=T r +(T i −T r )e kt
Substitute the known values:
95
70 + ( 180 − 70 ) e − 0.0904 ⋅ t 95=70+(180−70)e −0.0904⋅t
95
70 + 110 e − 0.0904 ⋅ t 95=70+110e −0.0904⋅t
25
110 e − 0.0904 ⋅ t 25=110e −0.0904⋅t
25 110
e − 0.0904 ⋅ t 110 25 =e −0.0904⋅t
5 22
e − 0.0904 ⋅ t 22 5 =e −0.0904⋅t
Take the natural logarithm ( ln ln) on both sides:
ln ( 5 22 )
− 0.0904 ⋅ t ln( 22 5 )=−0.0904⋅t t
ln ( 5 22 ) − 0.0904 t= −0.0904 ln( 22 5 )
Using a calculator:
ln ( 5 22 ) ≈ ln ( 0.2273 ) ≈ − 1.482 ln( 22 5 )≈ln(0.2273)≈−1.482 t
− 1.482 − 0.0904 ≈ 16.4 minutes t= −0.0904 −1.482 ≈16.4minutes Final Answers:
Temperature after 10 minutes: 114. 6 ∘ 114.6 ∘
Time to cool to 9 5 ∘ 95 ∘ : 16.4 minutes 16.4minutes
Solution
Here’s the cleaned-up and formatted version:
Newton’s Law of Cooling
The formula:
Where:
- : Temperature at time
- : Room temperature
- : Initial temperature
- : Cooling constant
- : Time
Given:
- Room temperature () =
- Initial temperature () =
- At minutes,
Step 1: Determine the cooling constant :
Using the formula:
Substitute the known values:
Simplify:
Take the natural logarithm () on both sides:
Solve for :
Using a calculator:
Thus:
Step 2a: Find the temperature after 10 minutes ():
Using the formula:
Substitute the known values:
Using a calculator:
Thus:
Step 2b: Find the time () for the temperature to cool to :
Using the formula:
Substitute the known values:
Take the natural logarithm () on both sides:
Solve for :
Using a calculator:
Final Answers:
- Temperature after 10 minutes:
- Time to cool to : minutes
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Math Problem Analysis
Mathematical Concepts
Exponential Functions
Logarithms
Newton's Law of Cooling
Formulas
T(t) = Tr + (Ti - Tr) e^(kt)
Theorems
Natural Logarithm Properties
Suitable Grade Level
Grades 10-12
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