Math Problem Statement

A thermometer reading 6degreesC is brought into a room with a constant temperature of 38 degrees C. If the thermometer reads 17degreesC after 2 ​minutes, what will it read after being in the room for 4 ​minutes? For 9 ​minutes? Question content area bottom Part 1 After 4 ​minutes, the thermometer will read

enter your response heredegreesC. ​(Do not round until the final answer. Then round to the nearest hundredth as​ needed.)

Solution

To solve this problem, we will use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula for Newton's Law of Cooling is:

T(t)=Tambient+(T0Tambient)ektT(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}}) e^{-kt}

where:

  • T(t)T(t) is the temperature of the object at time tt,
  • TambientT_{\text{ambient}} is the ambient (room) temperature,
  • T0T_0 is the initial temperature of the object,
  • kk is the cooling rate constant,
  • tt is the time elapsed.

Given:

  • Initial temperature, T0=6CT_0 = 6^\circ \text{C}
  • Ambient temperature, Tambient=38CT_{\text{ambient}} = 38^\circ \text{C}
  • Temperature after 2 minutes, T(2)=17CT(2) = 17^\circ \text{C}

Step 1: Find the cooling rate constant kk

Substitute the given values into the equation for T(2)T(2):

17=38+(638)e2k17 = 38 + (6 - 38) e^{-2k}

Simplify the equation:

17=3832e2k17 = 38 - 32 e^{-2k}

32e2k=2132 e^{-2k} = 21

e2k=2132e^{-2k} = \frac{21}{32}

Take the natural logarithm of both sides:

2k=ln(2132)-2k = \ln\left(\frac{21}{32}\right)

k=12ln(2132)k = -\frac{1}{2} \ln\left(\frac{21}{32}\right)

Calculating kk:

k=12ln(2132)0.207k = -\frac{1}{2} \ln\left(\frac{21}{32}\right) \approx 0.207

Step 2: Find the temperature after 4 minutes

Now, use the formula with t=4t = 4:

T(4)=38+(638)e0.2074T(4) = 38 + (6 - 38) e^{-0.207 \cdot 4}

T(4)=3832e0.828T(4) = 38 - 32 e^{-0.828}

Calculating e0.8280.437e^{-0.828} \approx 0.437:

T(4)=3832×0.437T(4) = 38 - 32 \times 0.437

T(4)=3813.98424.02T(4) = 38 - 13.984 \approx 24.02

So, after 4 minutes, the thermometer will read approximately 24.02°C.

Step 3: Find the temperature after 9 minutes

Repeat the process for t=9t = 9:

T(9)=38+(638)e0.2079T(9) = 38 + (6 - 38) e^{-0.207 \cdot 9}

T(9)=3832e1.863T(9) = 38 - 32 e^{-1.863}

Calculating e1.8630.155e^{-1.863} \approx 0.155:

T(9)=3832×0.155T(9) = 38 - 32 \times 0.155

T(9)=384.9633.04T(9) = 38 - 4.96 \approx 33.04

So, after 9 minutes, the thermometer will read approximately 33.04°C.

Would you like more details or have any questions?

Related Questions:

  1. What would the thermometer read after 5 minutes?
  2. How would the cooling rate change if the ambient temperature was different?
  3. What is the physical significance of the cooling rate constant kk?
  4. How does Newton's Law of Cooling apply to real-world situations?
  5. Can Newton's Law of Cooling be used to determine the time of death in forensic science?

Tip: When solving exponential equations like these, understanding logarithmic functions and their properties is essential for simplifying calculations effectively.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Newton's Law of Cooling

Formulas

T(t) = T_ambient + (T_0 - T_ambient) * e^(-kt)

Theorems

Newton's Law of Cooling

Suitable Grade Level

Grades 11-12