Master Specific Heat Calculations: Answers in SI Units

In summary: F" or do I need to use the whole equation?Your equations are correct, but your last sentence has it backwards. Sure you can. ΔF = 1.8 ΔC
  • #1
tj19926
8
0
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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  • #2
When you have a 1 °C change, what would be the corresponding change in °F? This conversion factor is independent of a specific temperature.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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}##.
 
Last edited:
  • #6
OP
Your first step is to convert all the units to SI units.
 
  • #7
DrClaude said:
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}##.

Your equations are correct, but your last sentence has it backwards.
 
  • #8
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
 
  • #9
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.
 

1. What is specific heat?

Specific heat is a physical property of a substance that measures the amount of heat required to raise the temperature of 1 gram of that substance by 1 degree Celsius.

2. How is specific heat different from heat capacity?

While specific heat measures the amount of heat needed to raise the temperature of a specific amount of a substance, heat capacity measures the amount of heat required to raise the temperature of an entire object.

3. Why is specific heat important?

Specific heat is important because it helps us understand how different substances respond to changes in temperature. It also plays a crucial role in many scientific and engineering applications, such as designing heating and cooling systems.

4. What factors affect the specific heat of a substance?

The specific heat of a substance is affected by its chemical composition, phase (solid, liquid, or gas), and temperature. It can also be influenced by external factors such as pressure and the presence of impurities.

5. How is specific heat measured?

Specific heat is typically measured using a calorimeter, which is an instrument that can accurately measure the heat absorbed or released by a substance during a temperature change. The specific heat is then calculated by dividing the amount of heat transferred by the change in temperature and the mass of the substance.

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