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Suppose 400.0 ) of heat transfer causes a sample's temperature to drop by 2.00°C. Calculate the heat capacity. AT= The algebraic signs are both As far as this sample is concerned, the process is the direction of heat flow is • the sample. We can rearrange the equation q= CAT to solve for C: C= AT • /K Note that the value of Cis the same whether we use J/C or J/K; we can also write the units as Jc or J K. Need help? Watch this. https://youtu.be/Lqb21Lt8 1w

Suppose 400.0 ) of heat transfer causes a sample's temperature to drop by 2.00°C. Calculate the heat capacity. AT= The algebraic signs are both As far as this sample is concerned, the process is the direction of heat flow is • the sample. We can rearrange the equation q= CAT to solve for C: C= AT • /K Note that the value of Cis the same whether we use J/C or J/K; we can also write the units as Jc or J K. Need help? Watch this. https://youtu.be/Lqb21Lt8 1w

Principles of Modern Chemistry
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter12: Thermodynamic Processes And Thermochemistry
Section: Chapter Questions
Problem 71AP
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Heat Capacity We define q, the amount of (energy transferred as) heat as directly proportional to the temperature change (AT): q-CAT where C is a positive constant (called the heat capacity of the sample). We can see from this equation that the larger the temperature change is (for a given sample), the more heat must have been transferred. If AT-1° C, then q and C are numerically equal; thus, heat capacity can be defined as the amount of heat needed to cause a temperature change of 1 : °C By convention AT is calculated by subtracting the initial temperature from the final temperature. AT = Tfinal - Tinitiat %3D Therefore, for a sample that warmed up, the algebraic sign of ATis positive • and the algebraic sign of q is positive For a sample that cooled down, the algebraic sign AT is negative and the algebraic sign of q is negative If the temperature of a sample decreased from 20.00°C to 18.00°C, then (in degrees Celsius) is AT - 18,00 : "c- 20.00 : °C = 2.00 : C In Kelvin, AT -( 18.00 • + 273.15) K-( 20.00 • + 273.15) K- 2.00 • K Notice that AT is the same in "C and K because the 273.15 that we add to the Celsius reading to convert to K cancels out when we subtract.
Transcribed Image Text:Heat Capacity We define q, the amount of (energy transferred as) heat as directly proportional to the temperature change (AT): q-CAT where C is a positive constant (called the heat capacity of the sample). We can see from this equation that the larger the temperature change is (for a given sample), the more heat must have been transferred. If AT-1° C, then q and C are numerically equal; thus, heat capacity can be defined as the amount of heat needed to cause a temperature change of 1 : °C By convention AT is calculated by subtracting the initial temperature from the final temperature. AT = Tfinal - Tinitiat %3D Therefore, for a sample that warmed up, the algebraic sign of ATis positive • and the algebraic sign of q is positive For a sample that cooled down, the algebraic sign AT is negative and the algebraic sign of q is negative If the temperature of a sample decreased from 20.00°C to 18.00°C, then (in degrees Celsius) is AT - 18,00 : "c- 20.00 : °C = 2.00 : C In Kelvin, AT -( 18.00 • + 273.15) K-( 20.00 • + 273.15) K- 2.00 • K Notice that AT is the same in "C and K because the 273.15 that we add to the Celsius reading to convert to K cancels out when we subtract.
Suppose 400.0 ) of heat transfer causes a sample's temperature to drop by 2.00°C. Calculate the heat capacity. AT= The algebraic signs are both As far as this sample is concerned, the process is the direction of heat flow is • the sample. We can rearrange the equation q= CAT to solve for C: C= AT • /K Note that the value of Cis the same whether we use J/C or J/K; we can also write the units as Jc or J K. Need help? Watch this. http://youtu.be/Lqb21L181w
Transcribed Image Text:Suppose 400.0 ) of heat transfer causes a sample's temperature to drop by 2.00°C. Calculate the heat capacity. AT= The algebraic signs are both As far as this sample is concerned, the process is the direction of heat flow is • the sample. We can rearrange the equation q= CAT to solve for C: C= AT • /K Note that the value of Cis the same whether we use J/C or J/K; we can also write the units as Jc or J K. Need help? Watch this. http://youtu.be/Lqb21L181w
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