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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Question
calculate for the specific heat with the given information
![Data:
Mcal
327.9g
Thot
100.0 oC
+ W
15.4g
Tcool = 20.1 oC
%3D
cal
312.5g
772.2g
Mw
Tfinal = 40.4 oC
M
sam
Analysis:
Finding the specific heat by the method of mixtures assumes that the energy lost by the cooling
metal is gained by the water as the mixture comes to thermal equilibrium. The energy gained by
the styrofoam cup is assumed to be negligible. Energy lost to the surroundings can also be
ignored if the experiment is carried out correctly and quickly.
Therefore: Quantity of energy lost by the sample = Quantity gained by water.
%3D
(Msam)(Csam)(Thot- Trinal) = (Mwater)(Cwater) (Tfinal – Tcool)
C.
water
= 1cal/go C
Use your data, algebra, and the above equation to find the specific heat of the sample in cal/go C.
Show your work below.
= [(Mwater) (1cal/go C) (T final
-T cold)]/ [(M am) (T hot -T final )] =
sam](https://content.bartleby.com/qna-images/question/0eb04277-ae49-49b6-9095-75ffb45bf9a2/167640ad-e297-4d43-b74e-f74dfeec75c3/g1w5ram_thumbnail.jpeg)
Transcribed Image Text:Data:
Mcal
327.9g
Thot
100.0 oC
+ W
15.4g
Tcool = 20.1 oC
%3D
cal
312.5g
772.2g
Mw
Tfinal = 40.4 oC
M
sam
Analysis:
Finding the specific heat by the method of mixtures assumes that the energy lost by the cooling
metal is gained by the water as the mixture comes to thermal equilibrium. The energy gained by
the styrofoam cup is assumed to be negligible. Energy lost to the surroundings can also be
ignored if the experiment is carried out correctly and quickly.
Therefore: Quantity of energy lost by the sample = Quantity gained by water.
%3D
(Msam)(Csam)(Thot- Trinal) = (Mwater)(Cwater) (Tfinal – Tcool)
C.
water
= 1cal/go C
Use your data, algebra, and the above equation to find the specific heat of the sample in cal/go C.
Show your work below.
= [(Mwater) (1cal/go C) (T final
-T cold)]/ [(M am) (T hot -T final )] =
sam
Expert Solution
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