A hot bar of iron at 1100 degrees Fahrenheit is quenched in oil at 350 degrees. After 2 minutes, the temperature of the iron is 750 degrees. (a) Find the value for k in Newton's Law of Cooling. Round your answer to the nearest thousandth. k = 0.315

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Chapter6: Exponential And Logarithmic Functions
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The answer for the first one is wrong it is not k=0.315

### Newton's Law of Cooling Example

A hot bar of iron at 1100 degrees Fahrenheit is quenched in oil at 350 degrees. After 2 minutes, the temperature of the iron is 750 degrees.

#### (a) Find the value for k in Newton's Law of Cooling. Round your answer to the nearest thousandth.
\[ k = 0.315 \]

#### (b) What will the temperature of the iron be after 10 minutes? Round your answer to the nearest degree.
\[ \text{Temperature} = 382 \text{ degrees Fahrenheit} \]

#### (c) How long will it take for the iron to reach 400 degrees? Round your answer to the nearest tenth of a minute.
\[ \text{Time} = 8.6 \text{ minutes} \]

**Question Help:** Video (link to the video or instructional material)

This exercise helps illustrate how to apply Newton's Law of Cooling to real-world problems involving temperature changes over time.
Transcribed Image Text:### Newton's Law of Cooling Example A hot bar of iron at 1100 degrees Fahrenheit is quenched in oil at 350 degrees. After 2 minutes, the temperature of the iron is 750 degrees. #### (a) Find the value for k in Newton's Law of Cooling. Round your answer to the nearest thousandth. \[ k = 0.315 \] #### (b) What will the temperature of the iron be after 10 minutes? Round your answer to the nearest degree. \[ \text{Temperature} = 382 \text{ degrees Fahrenheit} \] #### (c) How long will it take for the iron to reach 400 degrees? Round your answer to the nearest tenth of a minute. \[ \text{Time} = 8.6 \text{ minutes} \] **Question Help:** Video (link to the video or instructional material) This exercise helps illustrate how to apply Newton's Law of Cooling to real-world problems involving temperature changes over time.
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