# Specific Heat Question

have two specific heat questions that im 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 farenheit is the part that im having the hardest time with. I found that 1Kcal=3.968 BTU and i know how to convert to lbs, but i dont know how to work with the F since you cant simply use the 5/9ths formula

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

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

lewando
Homework Helper
Gold Member
When you have a 1 °C change, what would be the corresponding change in °F? This conversion factor is independent of a specific temperature.

Chestermiller
Mentor
have two specific heat questions that im 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 farenheit is the part that im having the hardest time with. I found that 1Kcal=3.968 BTU and i know how to convert to lbs, but i dont know how to work with the F since you cant simply use the 5/9ths formula

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

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

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

rude man
Homework Helper
Gold Member
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?

DrClaude
Mentor
Others have beat me to it, but here it is anyway...

The formula for converting a specific temperature from Fahrenheit to Celcius 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}##.

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Gold Member
OP
Your first step is to convert all the units to SI units.

Chestermiller
Mentor
Others have beat me to it, but here it is anyway...

The formula for converting a specific temperature from Fahrenheit to Celcius 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}##.

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 im 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

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)

Chestermiller
Mentor
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.

DrClaude
Mentor
Right. I'm getting tired .

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

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Chestermiller
Mentor
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

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?

Chestermiller
Mentor
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.