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The following lesson is one lecture in a series of Chemistry Programs developed by

The following lesson is one lecture in a series of Chemistry Programs developed by. Professor Larry Byrd Department of Chemistry Western Kentucky University. Temperature Scales. PART 3. EXAMPLE FIVE:. Convert 12.6  C into  F .

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The following lesson is one lecture in a series of Chemistry Programs developed by

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  1. The following lesson is one lecture in a series of Chemistry Programs developed by Professor Larry Byrd Department of Chemistry Western Kentucky University

  2. Temperature Scales PART 3

  3. EXAMPLE FIVE: Convert 12.6 C into F. 1. F = (1.8) (C) + 32Write the needed equation you had to learn 2. F = (1.8) (12.6) + 32Substitute the given value 3. F = 22.68 + 32Do math Round off 22.68 to the nearest tenth! 4. F = 22.7 + 32Do math 5. F = 54.7 6. Thus, 12.6 C = 54.7 F

  4. If we want to convert Fahrenheit readings into Celsius readings, we will need to rearrange the Fahrenheit (F) equation: 1. F = (1.8) (C) + 32Write the needed equation you had to learn 2. (1.8) (C) + 32 = FFirst, exchange sides 3. (1.8) (C) + 32 - 32 = F – 32Next, subtract 32 from both sides 4. (1.8) (C) + 32 - 32 = F - 32Do math 5. (1.8) (C) = F - 32

  5. If we want to convert Fahrenheit readings into Celsius readings, we will need to rearrange the Fahrenheit (F) equation: 6. Divide both sides by 1.8 7. Do math. Notice that the 1.8’s cancel 8. C =

  6. C = In words, the above equation states that to convert a value given in Fahrenheitinto a value in Celsius, we must do the following steps: 1. First subtract 32 from the Fahrenheit reading 2. Next, divide that value by 1.8 3. Then, round-off and give the new value with units of C Study the following examples:

  7. EXAMPLE SIX: The oral reading for a normal body temperature of a patient is 98.6 F. Convert that value into a Celsius reading. Notice, that we must round-off to the nearest tenth of a degree. 1. C =Write needed equation 2. C = Substitute the given value 3. C =Do division and round-off to the nearest tenth 4. C =37.0 Give final answer to the nearest tenth, since 98.6 F was to the nearest tenth Thus, 98.6 °F is equal to 37.0 °C

  8. EXAMPLE SEVEN: Convert-16.FintoC: 1. C =Write the needed equation • C =Substitute the given value and • Do algebraically 3. C =Watch the negative sign! Divide ! This answer must be rounded off to the nearest whole number since the starting temperature, -16 F , was a whole number 4. C = - 26.6666666 5. C = - 27. C 6.Thus, - 16. F is equal to -27. C

  9. EXAMPLE EIGHT: Convert 16.69 F into C. [ Round-off to the nearest hundredth of a degree ] 1. C =Write the needed equation 2. C =Substitute the given value. 3. C =Do math , watch signs and Do algebraically! 4. C = - 8.505555Round-off to the nearest hundredth 5. C = - 8.51 6. Thus, 16.69 F is the equal to -8.51 C.

  10. EXAMPLE NINE: Convert 75.6 F into K In doing this conversion, we must first convert F into C and then convert the Celsius( C ) value into a Kelvin Value. (A) Convert 75.6 F into C (Starting temperature is to the nearest tenth) 1. C =Write the needed equation 2. C =Substitute the given value and Do algebraically 3. C =Do Math! 4. C = 24.22222Round-off to the nearest tenth 5. C = 24.2 6. Thus, 75.6 F is equal to 24.2 C.

  11. EXAMPLE NINE: Convert 75.6 F into K In doing this conversion, we must first convert F into C and then convert the Celsius( C ) value into a Kelvin Value. (B) "NEXT" Convert 24.2 C into Kelvin: 1. K = C + 273.2Write the needed equation 2. K = 24.2 + 273.2Substitute the given value Notice, that the answer was rounded-off to the nearest tenth 3. K = 297.4 Thus, 75.6 F is equal to 24.2 C and is also equal to 297.4 K

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