How would a samples specific heat capacity be affected if its temperature was lowered before entering a calorimeter?

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greym
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How would a samples specific heat capacity be affected if its temperature was lowered before entering a calorimeter?

Post by greym »

I am conducting a lab where a metal sample was heated in a hot water bath, and transferred into a Styrofoam calorimeter filled with room temp water. The thermal equilibrium temperature was recorded, while the initial temperature of the metal was recorded inside the hot water bath and the maters initial temperature was measured prior to the metals entry. We had to use forceps to move the metal from bath to calorimeter, and due to that some heat was lost mid-transfer. How would this effect the variables of a specific heat capacity equation of:

(Cm = Mw x Cw x ΔTw / -Mm x ΔTm ) where M is mass, C is heat capacity, ΔT is temperature change, and m and w are used to state what that variable is describing, m being the sample and w being the water.

I would appreciate a reply as quick as possible.

-Grey
CarolOKlaNOLA
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Re: How would a samples specific heat capacity be affected if its temperature was lowered before entering a calorimeter?

Post by CarolOKlaNOLA »

Exactly WHAT are you trying to measure? The specific heat capacity of an unknown metal? In addition to measuring the heat capacity, are you also trying to determine what the metal is? Lowering the temperature of the metal during the transfer to the calorimeter is not going to change the specific heat capacity of the metal, because you are measuring the CHANGE in the temperature of of the metal as it cools off. The specific heat capacity of the metal will not change. Were the calipers insulated in any way?. What were the calipers made of. All of that should go into the into the description of the experiment that i assume you will be turning in. Your formula doesn't even include the for specific heat capacity and is incorrect.. isn't specific heat capacity what you are trying to measure.

The formula is
Q = mc∆t

Q = energy added or lost in joules
m = mass of the water
c= heat capacity of the metal
∆T = the difference in temperature of the metal in Celsius or Kelvin.

The calorimeter measures the heat energy transferred. The unknown variable here is the heat capacity of the metal. You know the mass of the water, and the difference in temperature of the matter, Since you have three of the four quantities in the formula, calculating the heat specific heat capacity of the metal should be relatively easy. The experiment should be repeated three times so you can confirm or deny the validity of your results statistically. IF this is an assignment, you need to talk to your teacher about what the teacher wants to be reported. Even though this may seem like a relatively simple experiment that has been done by millions of students, even simple experiments can be VERY frustrating because human beings beings make mistakes so often. Learning from your mistakes is what LIFE and education is all about. life is continuing education process.

Source for the formula:
https://www.google.com/search?q=specifi ... nw9YF4M%3A
greym
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Re: How would a samples specific heat capacity be affected if its temperature was lowered before entering a calorimeter?

Post by greym »

We were taught that since -Q(heat lost) = Q(heat gained) we can substitute each value of Q for the equation cmΔT where heat lost is relating to the metal sample and heat gained relates to the water. By recording the metals mass and its temperature change, along with waters mass and temperature change, and using waters theoretical specific heat capacity, we can get an experimental specific heat capacity for the metal.
−(Specific Heat Capacity of Metal) · (Mass Of Metal) · (Temperature Change of Metal)

is equal to

(Specific Heat Capacity of Water) · (Mass Of Water) · (Temperature Change of Water)
Another factor is Temperature change, which can be expressed as:
Final Temperature(Thermal Equilibrium Temp) − Initial Temperature
What I am trying to do is determine how heat being lost mid-transfer would affect the experimental specific heat capacity of the metal if the metals initial temperature was recorded in the hot water bath.
kmyers2040
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Re: How would a samples specific heat capacity be affected if its temperature was lowered before entering a calorimeter?

Post by kmyers2040 »

Hi Grey,
Your equations are all correct and you are using the heat equilibirium equations to describe the heat transfer process occurring in the calorimeter: heat is lost from the metal and gained by the water in the calorimeter.
In your experiment you measured the metal's intial temperature in the hot water bath and you are assuming that is the metal's intial temperature when it is placed in the room temparature water in the calorimeter. But as you have correctly pointed out, there is actually some heat loss during the transfer process. That means the metal will have gotten to a lower temperature before it is placed in the room temperature water.
So where it affects your experiment calculation is the temperature change for the metal.
Since your measured intitial temperature value of the metal (measured in the hot water bath )is higher than the actual temperature the metal was at (after transfer heat loss), your calculated temperature change would be greater than the actual temperature change. And when you plug your calculated temperature change into the specific heat capacity equation, your calculated experimental specific heat capacity would be smaller than the actual specific heat capacity.
Hopefully that all made sense, post back if you have more questions.
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