Converting Moon Temperatures to Celsius and Fahrenheit

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SUMMARY

The discussion focuses on converting lunar surface temperatures from Kelvin to Celsius and Fahrenheit. The correct conversion formulas are established: Celsius (Tc) to Fahrenheit (Tf) is given by Tf = (9/5) * Tc + 32. The temperatures on the moon are 371 K (97.85 °C) during the day and 1.09 x 10² K (-164.15 °C) at night. The key takeaway is the importance of accounting for the freezing point differences between Celsius and Fahrenheit when performing conversions.

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Homework Statement



On the moon the surface temperature ranges from 371 K during the day to 1.09 102 K at night.
(a) What are these temperatures on the Celsius scale?
(b) What are these temperatures on the Fahrenheit scale?

Homework Equations


T=Tc+273.15 T=Kelvin temperature Tc=celsius
9/5 F degrees/(1 C)

The Attempt at a Solution



I got both answers right for part a but for part b the answer solution says that my answers are incorrect.

What I did was first plugged in each temperature given in the problem separately into the equation to find celsius. I ended up with 97.85 C and -164.15 C which were correct! Then I used the second conversion 9/5 F degrees/(1 C) to find each of the temperature for farenheight using the celsius degrees answers from part a I got 176.13 degrees F and -295.15.
I think I worked this out correctly but the key says I'm wrong
 
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There's an offset in the F temperature scale. Water freezes at 0C, but not at 0F. Does that help?
 
You're forgetting that the zero points of the Fahrenheit and Celsius scales are not the same. More specifically, 0 degrees C is equal to 32 degrees F. As a result, you have to shift. I'll leave to you to determine whether to shift, then convert, or whether to convert, and then shift.
 
Oops

cepheid said:
You're forgetting that the zero points of the Fahrenheit and Celsius scales are not the same. More specifically, 0 degrees C is equal to 32 degrees F. As a result, you have to shift. I'll leave to you to determine whether to shift, then convert, or whether to convert, and then shift.
So I need to subtract 32 degrees from each answer to take an account for the freezing pt.
 
scw287 said:
So I need to subtract 32 degrees from each answer to take an account for the freezing pt.

Not quite.

97.85 C means 97.85 Celsius degrees above freezing. When you convert these to Fahrenheit degrees, it means 176.13 Fahrenheit degrees above freezing, where freezing is 32 F. So...
 
first i would take the celsius and subtract 32 degrees since its above the farenheight freezing pt. and then convert to farenheight by multiplying by (9/5)...right...and in the other part i would add 32 degrees since its below the zero pt??
 
Both scales are equivalent at -40 degrees. Start from there and use the ratio you know. Be sure to readjust your answer to account for the 40 degrees.
 
scw287 said:
first i would take the celsius and subtract 32 degrees since its above the farenheight freezing pt. and then convert to farenheight by multiplying by (9/5)...right...and in the other part i would add 32 degrees since its below the zero pt??

No and no. The conversion algorithm is always the same, and you have got it wrong. Think about it this way: How would you convert 0C to Fahrenheit?

EDIT: By the way, I was hoping you'd complete the thought here, that was the point of my post:

cepheid said:
97.85 C means 97.85 Celsius degrees above freezing. When you convert these to Fahrenheit degrees, it means 176.13 Fahrenheit degrees above freezing, where freezing is 32 F. So...
 
Last edited:
Man I was making this problem much tougher than it was! thanks for your help!
 
  • #10
o sorry, When you convert these to Fahrenheit degrees, it means 176.13 Fahrenheit degrees above freezing, where freezing is 32 F. so, Tf = (9/5)*Tc+32
 
  • #11
scw287 said:
o sorry, When you convert these to Fahrenheit degrees, it means 176.13 Fahrenheit degrees above freezing, where freezing is 32 F. so, Tf = (9/5)*Tc+32

This is the exact point I was getting at, and the equation you posted above is the correct conversion formula! I'm glad you arrived at it by reasoning it out, that's what I was hoping to coax you to do! :smile:
 
  • #12
sometimes i just try to complicate problems...thanks for your help I appreciate it very much ;-)
 

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