Why Must a Calorimeter Be Saturated with Water Vapor?

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Homework Help Overview

The discussion revolves around the use of a bomb calorimeter to measure heat output during the combustion of methanol and the calculation of the calorimeter's heat capacity. Participants are exploring the implications of the calorimeter being saturated with water vapor in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the heat capacity using the enthalpy of combustion and temperature change but expresses confusion regarding the negative value obtained. Some participants clarify the interpretation of heat capacity and suggest using the positive value of heat for calculations.

Discussion Status

Participants are actively discussing the correct approach to calculating heat capacity, with some providing clarifications on the definitions and equations involved. There is a mix of interpretations regarding the signs used in calculations, but no explicit consensus has been reached.

Contextual Notes

There is mention of the need for the calorimeter to be saturated with water vapor, although the specific reasons for this requirement are not detailed in the discussion.

Armitage12
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Homework Statement


A bomb calorimeter is used to measure the overall heat output. It is calibrated by burning 1.00g of methanol (Change in enthalpy of combustion- 715 kJ mol–1) in O2 which produces a temperature rise of 8.40 K. Use this information to determine the heat capacity of the calorimeter. Why must the calorimeter be saturated with water vapour?

Going over past papers and seen this question. Just a little confused as to how to do it. Previous capacity questions I've done, have given the power and time, which i multipled for q and and divided by the change in temperature for the heat capacity.

I understand the to get q= change in combustion x no.of moles
so -22.34

Then plug it into Cv= q/Change in temp

to get the heat capacity of the calorimeter at -2.66 KJ/K

But it doesn't make sense to me? isn't the definition of heat capacity, the amount of energy needed to raise the temp by 1k?
So a negative answer can't be right?
 
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The 715 is the amount of heat that would have had to be removed to keep the temperature constant (that's where the minus sign comes from). It wasn't removed, but, instead, went into raising the temperature of the calorimeter. So the calorimeter received the 715. The heat capacity is positive.

Chet
 
So do I use the same equation but instead of -715 the positive value 715?
So.. multiply 715 by the moles of methanol then divide by the 8.4k temp rise? to get 2.2kJ/K?
 
Armitage12 said:
So do I use the same equation but instead of -715 the positive value 715?
Yes.
So.. multiply 715 by the moles of methanol then divide by the 8.4k temp rise? to get 2.2kJ/K?
I get 2.66, in agreement with your original answer.

Chet
 
Last edited:
Hey, also helpful to note that the equations are as follows:

Heat Capacity (Amount of heat needed to raise the temperature of an object by one degree) : C = Q/∆T Where Q is Heat Energy and T is temperature.

Specific Heat Capacity (Amount of heat needed to raise the temperature of 1kg of an object by 1 degree) : c = Q/m∆T where Q is Heat Energy, m is mass and T is temperature.

Also keep in mind temperature is in degrees Celsius and ALWAYS absolute value. Don't write it as negative.
 
tashad said:
Hey, also helpful to note that the equations are as follows:

Heat Capacity (Amount of heat needed to raise the temperature of an object by one degree) : C = Q/∆T Where Q is Heat Energy and T is temperature.

Specific Heat Capacity (Amount of heat needed to raise the temperature of 1kg of an object by 1 degree) : c = Q/m∆T where Q is Heat Energy, m is mass and T is temperature.

Also keep in mind temperature is in degrees Celsius and ALWAYS absolute value. Don't write it as negative.
Hi Tashad. Welcome to Physics Forums.

Did you really think that Armitage12 did not already know these things?

Chet
 
Last edited:
Chestermiller said:
Hi Tashad. Welcome to Physics Forums.

Did you really think that Armitage12 did not already know these things?

Chet

Hi Chet. Thanks for the kind welcome. I just browsed to the "Introductory Physics" section and assumed its where all the simple questions would be asked. My apologies.
 

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