Master Specific Heat Calculations: Answers in SI Units

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Homework Help Overview

The discussion revolves around specific heat calculations, particularly focusing on the conversion of specific heat values and temperature changes between Celsius and Fahrenheit. The original poster presents two questions related to specific heat in different units and the energy required to change the temperature of a large body of water, specifically the Pacific Ocean.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the conversion of specific heat from calories to BTUs and question how to handle temperature changes in Fahrenheit versus Celsius. There are discussions about the relationship between temperature intervals in different units and the implications for specific heat calculations.

Discussion Status

Participants have provided various insights into the conversion factors between Celsius and Fahrenheit, with some suggesting that the original poster should convert all units to SI. There is an ongoing exploration of how to apply these conversions to the specific heat questions posed, with multiple interpretations of the relationships between the units being discussed.

Contextual Notes

Some participants note the need for clarity regarding the volume and density of the Pacific Ocean, as well as the significance of significant figures in the final answers. There is also mention of the original poster's confusion regarding the use of Fahrenheit in the calculations.

tj19926
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have two specific heat questions that I am having trouble with.
Though i thought this was a chemistry topic, we are learning it in my physics class as well


1. The specific heat of aluminum is 0.22 cal/g *C

What is the specific heat in Btu/lb*F

The Fahrenheit is the part that I am having the hardest time with. I found that 1Kcal=3.968 BTU and i know how to convert to lbs, but i don't know how to work with the F since you can't simply use the 5/9ths formula



2. How many calories would it take to raise the temperature of the entire Pacific Ocean by 1*F.

Once again, I am having the most trouble with the Fahrenheit part. I found that the volume of the pacific ocean is 4.14x10^8 Km3 and the density is 1027 Kg/m^3.





Answers need to be in SI. Thanks for your help!
 
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When you have a 1 °C change, what would be the corresponding change in °F? This conversion factor is independent of a specific temperature.
 
tj19926 said:
have two specific heat questions that I am having trouble with.
Though i thought this was a chemistry topic, we are learning it in my physics class as well


1. The specific heat of aluminum is 0.22 cal/g *C

What is the specific heat in Btu/lb*F

The Fahrenheit is the part that I am having the hardest time with. I found that 1Kcal=3.968 BTU and i know how to convert to lbs, but i don't know how to work with the F since you can't simply use the 5/9ths formula

Sure you can. ΔF = 1.8 ΔC
2. How many calories would it take to raise the temperature of the entire Pacific Ocean by 1*F.

Once again, I am having the most trouble with the Fahrenheit part. I found that the volume of the pacific ocean is 4.14x10^8 Km3 and the density is 1027 Kg/m^3.


Answers need to be in SI. Thanks for your help!
What is the heat capacity of sea water?
How many degrees C is 1 degree F?
 
Think about what "BTU/lb-F" means. It means it takes Q BTU to heat 1 lb of aluminum 1 deg F.
Now, 1 deg C is more than 1 deg. F. So would it take more or less than Q BTU to heat 1 lb Al 1 deg C?
If so, by how much?
 
Others have beat me to it, but here it is anyway...

The formula for converting a specific temperature from Fahrenheit to celsius is
$$
\frac{5}{9}(^\circ \mathrm{F} - 32) = {}^\circ \mathrm{C}
$$
What you are deling with here are intervals between two temperatures. You therefore have
$$
\begin{align}
\Delta^\circ \mathrm{C} &= {}^\circ \mathrm{C}_2 - {}^\circ \mathrm{C}_1\\
&= \frac{5}{9}(^\circ \mathrm{F}_2 - 32) - \frac{5}{9}(^\circ \mathrm{F}_1 - 32)\\
&= \frac{5}{9}(^\circ \mathrm{F}_2 - {}^\circ \mathrm{F}_1 ) \\
&= \frac{5}{9} \Delta^\circ \mathrm{F}
\end{align}
$$
[STRIKE]As you see, you need to use the factor ##5/9## only. A change of ##1^\circ \mathrm{C}## is equal to a change of ##(5/9) ^\circ \mathrm{F}##, and a change of ##1^\circ \mathrm{F}## is the same as a change of ##(9/5) ^\circ \mathrm{C}##.[/STRIKE]
As you see, you need to use the factor ##5/9## only. A change of ##1^\circ \mathrm{C}## is equal to a change of ##(9/5) ^\circ \mathrm{F}##, and a change of ##1^\circ \mathrm{F}## is the same as a change of ##(5/9) ^\circ \mathrm{C}##.
 
Last edited:
OP
Your first step is to convert all the units to SI units.
 
DrClaude said:
Others have beat me to it, but here it is anyway...

The formula for converting a specific temperature from Fahrenheit to celsius is
$$
\frac{5}{9}(^\circ \mathrm{F} - 32) = {}^\circ \mathrm{C}
$$
What you are deling with here are intervals between two temperatures. You therefore have
$$
\begin{align}
\Delta^\circ \mathrm{C} &= {}^\circ \mathrm{C}_2 - {}^\circ \mathrm{C}_1\\
&= \frac{5}{9}(^\circ \mathrm{F}_2 - 32) - \frac{5}{9}(^\circ \mathrm{F}_1 - 32)\\
&= \frac{5}{9}(^\circ \mathrm{F}_2 - {}^\circ \mathrm{F}_1 ) \\
&= \frac{5}{9} \Delta^\circ \mathrm{F}
\end{align}
$$
As you see, you need to use the factor ##5/9## only. A change of ##1^\circ \mathrm{C}## is equal to a change of ##(5/9) ^\circ \mathrm{F}##, and a change of ##1^\circ \mathrm{F}## is the same as a change of ##(9/5) ^\circ \mathrm{C}##.

Your equations are correct, but your last sentence has it backwards.
 
Chestermiller said:
Sure you can. ΔF = 1.8 ΔC

What is the heat capacity of sea water?
How many degrees C is 1 degree F?

so for the first problem, i got it to 0.396 Btu/lb*C
From that point I am still not understanding how to implement that equation to get it to F


and for the second problem I know i need to use Q=mcDeltaT
Q=(7.33x10^14kg kg of water)(3850 J/kg*C)(0.56*c)

Is that right prior to converting to calories? And also with sig figs, how many would have to be in the answer? would it just be one since the question says "to 1°F?

Thanks
 
rude man said:
Think about what "BTU/lb-F" means. It means it takes Q BTU to heat 1 lb of aluminum 1 deg F.
Now, 1 deg C is more than 1 deg. F. So would it take more or less than Q BTU to heat 1 lb Al 1 deg C?
If so, by how much?

I think it would take more. I'm still not sure how to work in a conversion factor though. I worked the problem up to 0.396 Btu/lb*C (which i think is correct)
 
  • #10
tj19926 said:
I think it would take more. I'm still not sure how to work in a conversion factor though. I worked the problem up to 0.396 Btu/lb*C (which i think is correct)

A BTU is the amount of heat required to raise the temperature of 1 lb of water 1 degree F.
A calorie is the amount of heat required to raise the temperature of 1 gm of water 1 degree C.

So, 1BTU/lb-F is exactly the same as 1cal/gmC. The conversion factor is 1

Show us how you calculated the mass of the Pacific ocean. Your result doesn't look right. Otherwise, your calculation for part 2 looks OK.
 
  • #11
Chestermiller said:
Your equations are correct, but your last sentence has it backwards.
Right. I'm getting tired :redface:.
 
  • #12
Chestermiller said:
A BTU is the amount of heat required to raise the temperature of 1 lb of water 1 degree F.
A calorie is the amount of heat required to raise the temperature of 1 gm of water 1 degree C.

So, 1BTU/lb-F is exactly the same as 1cal/gmC. The conversion factor is 1

Show us how you calculated the mass of the Pacific ocean. Your result doesn't look right. Otherwise, your calculation for part 2 looks OK.

I looked it up and it said the ocean's volume was 7.14x10^8Km^3
so i multiplied this x 1027kg/m^3(density of sea water) then x 1000m^3/1km^3
 
Last edited:
  • #13
tj19926 said:
I looked it up and it said the ocean's volume was 7.14x10^18Km^3
so i multiplied this x 1027kg/m^3(density of sea water) then x 1000m^3/1km^3

That should be 1000000000m^3/1km^3
 
  • #14
Chestermiller said:
That should be 1000000000m^3/1km^3



thank you very much :) Also, if i need to answer this problem to the correct number of sig figs, how many sig figs would that be?
 
  • #15
tj19926 said:
thank you very much :) Also, if i need to answer this problem to the correct number of sig figs, how many sig figs would that be?
In my judgement, 3 significant figures.
 

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