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:
T(t)=Tr+(TiTr)ektT(t) = T_r + (T_i - T_r) e^{kt}

Where:

  • T(t)T(t): Temperature at time tt
  • TrT_r: Room temperature
  • TiT_i: Initial temperature
  • kk: Cooling constant
  • tt: Time

Given:

  • Room temperature (TrT_r) = 7070^\circ
  • Initial temperature (TiT_i) = 180180^\circ
  • At t=5t = 5 minutes, T(5)=140T(5) = 140^\circ

Step 1: Determine the cooling constant kk:

Using the formula:
T(t)=Tr+(TiTr)ektT(t) = T_r + (T_i - T_r) e^{kt}

Substitute the known values:
140=70+(18070)e5k140 = 70 + (180 - 70) e^{5k}

Simplify:
140=70+110e5k140 = 70 + 110 e^{5k}
70=110e5k70 = 110 e^{5k}
70110=e5k\frac{70}{110} = e^{5k}
711=e5k\frac{7}{11} = e^{5k}

Take the natural logarithm (ln\ln) on both sides:
ln(711)=5k\ln\left(\frac{7}{11}\right) = 5k

Solve for kk:
k=ln(711)5k = \frac{\ln\left(\frac{7}{11}\right)}{5}

Using a calculator:
kln(0.6364)50.45250.0904k \approx \frac{\ln(0.6364)}{5} \approx \frac{-0.452}{5} \approx -0.0904

Thus:
k0.0904k \approx -0.0904


Step 2a: Find the temperature after 10 minutes (t=10t = 10):

Using the formula:
T(t)=Tr+(TiTr)ektT(t) = T_r + (T_i - T_r) e^{kt}

Substitute the known values:
T(10)=70+(18070)e0.090410T(10) = 70 + (180 - 70) e^{-0.0904 \cdot 10}
T(10)=70+110e0.904T(10) = 70 + 110 e^{-0.904}

Using a calculator:
e0.9040.405e^{-0.904} \approx 0.405
T(10)=70+110(0.405)T(10) = 70 + 110(0.405)
T(10)70+44.55114.55T(10) \approx 70 + 44.55 \approx 114.55^\circ

Thus:
T(10)114.6T(10) \approx 114.6^\circ


Step 2b: Find the time (tt) for the temperature to cool to 9595^\circ:

Using the formula:
T(t)=Tr+(TiTr)ektT(t) = T_r + (T_i - T_r) e^{kt}

Substitute the known values:
95=70+(18070)e0.0904t95 = 70 + (180 - 70) e^{-0.0904 \cdot t}
95=70+110e0.0904t95 = 70 + 110 e^{-0.0904 \cdot t}
25=110e0.0904t25 = 110 e^{-0.0904 \cdot t}
25110=e0.0904t\frac{25}{110} = e^{-0.0904 \cdot t}
522=e0.0904t\frac{5}{22} = e^{-0.0904 \cdot t}

Take the natural logarithm (ln\ln) on both sides:
ln(522)=0.0904t\ln\left(\frac{5}{22}\right) = -0.0904 \cdot t

Solve for tt:
t=ln(522)0.0904t = \frac{\ln\left(\frac{5}{22}\right)}{-0.0904}

Using a calculator:
ln(522)ln(0.2273)1.482\ln\left(\frac{5}{22}\right) \approx \ln(0.2273) \approx -1.482
t1.4820.090416.4minutest \approx \frac{-1.482}{-0.0904} \approx 16.4 \, \text{minutes}


Final Answers:

  1. Temperature after 10 minutes: 114.6114.6^\circ
  2. Time to cool to 9595^\circ: 16.416.4 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