Heat CapacityWe define q, the amount of (energy transferred as) heat as directly proportional to the temperature change (AT):q-CATwhere 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 canbe defined as the amount of heat needed to cause a temperature change of 1 : °CBy convention AT is calculated by subtracting the initial temperature from the final temperature.AT = Tfinal - Tinitiat%3DTherefore, for a sample that warmed up, the algebraic sign of ATis positive • and the algebraic sign of q is positiveFor a samplethat cooled down, the algebraic sign AT is negative and the algebraic sign of q is negativeIf the temperature of a sample decreased from 20.00°C to 18.00°C, then (in degrees Celsius) isAT - 18,00 : "c- 20.00 : °C = 2.00 : CIn Kelvin,AT -( 18.00 • + 273.15) K-( 20.00 • + 273.15) K- 2.00 • KNotice 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 bothAs far as this sample is concerned, the process isthe direction of heat flow is• the sample.We can rearrange the equation q= CAT to solve for C:C=AT• /KNote 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

Answered: Image /qna-images/answer/fe4d04fc-37b8-4786-a671-0af87dcb287b.jpg

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

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Thermochemistry

Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.

Exergonic Reaction

The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.

Question

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

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