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Chapter 13: Problem 4

(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?

Short Answer

Expert verified
Is there a temperature at which the numerical value of kelvins becomes equal to the numerical value of Fahrenheit degrees? Solution: The numerical values of Celsius and Fahrenheit degrees become equal at -40 degrees Celsius (-40掳F). There is no temperature at which the numerical value of kelvins is equal to the numerical value of Fahrenheit degrees, as the Kelvin scale does not have negative values.

Step by step solution

01

Part a: Find the temperature at which Celsius equals Fahrenheit

To find the temperature at which the Celsius and Fahrenheit scales are equal, we need to solve the equation: C = (F - 32) * (5/9) Rearrange the equation to solve for F: F = (9/5)C + 32 Since we're looking for the temperature at which C and F are numerically equal, we can substitute C for F in the equation: C = (9/5)C + 32 Now, we need to solve for C:
02

Subtract C from both sides

Subtract C from both sides of the equation: 0 = (4/5)C + 32
03

Isolate C

To isolate C, we need to get rid of the constant term, which is 32. We can do this by subtracting 32 from both sides of the equation: -32 = (4/5)C
04

Solve for C

To solve for C, we can multiply both sides of the equation by the reciprocal of the coefficient of C, which is (5/4): C = -40 So at -40 degrees Celsius, the numerical value of Celsius degrees equals the numerical value of Fahrenheit degrees.
05

Part b: Find the temperature at which Kelvin equals Fahrenheit

To find the temperature at which the Kelvin and Fahrenheit scales are equal, we need to solve the equation: K = (F - 32) * (5/9) + 273.15 Rearrange the equation to solve for F: F = (9/5)(K - 273.15) + 32 Since we're looking for the temperature at which K and F are numerically equal, we can substitute K for F in the equation: K = (9/5)(K - 273.15) + 32 Now, we need to solve for K:
06

Subtract K from both sides

Subtract K from both sides of the equation: 0 = (-4/5)K + (9/5)(-273.15) + 32
07

Solve for K

To solve for K, isolate the term with K by adding (4/5)K to both sides of the equation: (4/5)K = (9/5)(-273.15) + 32 Now, multiply both sides of the equation by the reciprocal of the coefficient of K, which is (5/4): K = -229.73 It is important to note that the result we got is not valid because the Kelvin scale starts at 0 and does not have negative values. Therefore, there is no temperature at which the numerical value of kelvins is equal to the numerical value of Fahrenheit degrees.

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Most popular questions from this chapter

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