Calculating F using Sig Figs | (5/9) (6.5) + 32

[Ognerok]

Homework Statement

I need to calculate F = (5/9) (6.5) + 32 using the correct number of sig figs.

Homework Equations

F = (5/9) (c) + 32

The Attempt at a Solution

Got down to 3.6 + 32; should it just be 35.6 or 36?

 

[Feldoh]

3.6 + 32 you're adding a number with two sig figs to another number with two sig figs, so would you have two sig figs (36) or three (35.6) in the final answer?

 

[Ognerok]

3.6 + 32 you're adding a number with two sig figs to another number with two sig figs, so would you have two sig figs (36) or three (35.6) in the final answer?

Ah, nevermind. Considering 32 is an exact number anyway (a "counting" number) in which sig figs aren't counted in the 32.

I guess if it was 3.6 + 32.0000, then sig figs would be counted for the 32.0000 (which is, 4 decimal places vs. 1 decimal place; result should have 1 decimal place).

 

[Chi Meson]

No.

When adding and subtracting, you do not count the number of digits, you find the least precise decimal position.

The 9/5 (assuming this is not exact, coming out of a specific ratio in a formula) will cause the result of (9/5)(6.5) to have a single digit (following rules of multiplication) or 4.

4 is significant to the "unit" (or the "ones") which gets added to 32 which is also significant to the "unit." They are added, and the sum will be significant to the unit. Answer is 36.

Anyway, unless you were told specifically that 32 was a "counted" number, and therefore exact, you cannot make that assumption. And since the equation implies you are calculating a force, I'm not sure how you can be so sure about counting exactly 32 individual Newtons of force (all lined up in the same direction?)

Furthermore, at some point even counting becomes erroneous. At what point? count out 100 pennies; are you sure you have 100? How about after counting 1000? 10,000? When does the possibility of a miscount become "highly probable"?

If this is an equation for Force (implied by the "F"), then the 32 must be a force which means it must have been measured. All measurements are inherently flawed and suffer from finite precision and some inaccuracy. 32 is a two sig measurement.

 

[kuruman]

...

If this is an equation for Force (implied by the "F"), then the 32 must be a force which means it must have been measured. All measurements are inherently flawed and suffer from finite precision and some inaccuracy. 32 is a two sig measurement.

For whatever it's worth, looks like he is converting temperature from Fahrenheit to Celsius.

** Edit **
I meant from Celsius to Fahrenheit.

 

[Redbelly98]

Ah, nevermind. Considering 32 is an exact number anyway (a "counting" number) in which sig figs aren't counted in the 32.

I guess if it was 3.6 + 32.0000, then sig figs would be counted for the 32.0000 (which is, 4 decimal places vs. 1 decimal place; result should have 1 decimal place).

Looks good to me. For what it's worth, we are dealing with number of places past the decimal rather than significant figures, but it looks like you probably realize that.

 

[Chi Meson]

For whatever it's worth, looks like he is converting temperature from Fahrenheit to Celsius.

** Edit **
I meant from Celsius to Fahrenheit.

I was very tired last night. Still, I cannot believe I didn't recognize the c=>F formula. In this case, the 9/5 is an exact ratio and doesn't change the number of sig figs in (5/9)x 6.5, thus 3.6.

And since both 0˚C and 32˚F are defined as the same temperature, the 32 is also absolute and, when adding, does not change the "least precise" position of the answer.

That means the answer is 35.6˚F.