extrapolation of a curve. 2. A student determined the calorimeter constant of the calorimeter, using the procedure described in this module. The student added 50.00 mL of cold water to 50.00 mL of heated, distilled water in a Styrofoam cup. The initial temperature of the cold water was 21.00 °C and of the hot water, 29.15 °C. The maximum temperature of the mixture was found to be 24.81 °C. Assume the density of water is 1.00 g mL-1 and the specific heat is 4.184 J g¯1K-!. bollos (1) Determine the AT for the hot water and the cold water. amil At cold water = 24.81-21.00 = 3.81°Cos nim AT hot = 24,81- 29.15 = -4.34°C 2.0 121 0.1 EP SS If 28.SS Call 08 TI a.1 ar anbim hot water - 4.34°C PESS 81 answer cold water 381°C dogon Loximnu sdt lo anutSTeqmd mom adt ontmolab bne sisb ami answer (2) Calculate the heat lost by the hot water. 4.34°C answer Aqg slt moth (3) Calculate the heat gained by the cold water. 3.31OC answer

extrapolation of a curve. 2. A student determined the calorimeter constant of the calorimeter, using the procedure described in this module. The student added 50.00 mL of cold water to 50.00 mL of heated, distilled water in a Styrofoam cup. The initial temperature of the cold water was 21.00 °C and of the hot water, 29.15 °C. The maximum temperature of the mixture was found to be 24.81 °C. Assume the density of water is 1.00 g mL-1 and the specific heat is 4.184 J g¯1K-!. bollos (1) Determine the AT for the hot water and the cold water. amil At cold water = 24.81-21.00 = 3.81°Cos nim AT hot = 24,81- 29.15 = -4.34°C 2.0 121 0.1 EP SS If 28.SS Call 08 TI a.1 ar anbim hot water - 4.34°C PESS 81 answer cold water 381°C dogon Loximnu sdt lo anutSTeqmd mom adt ontmolab bne sisb ami answer (2) Calculate the heat lost by the hot water. 4.34°C answer Aqg slt moth (3) Calculate the heat gained by the cold water. 3.31OC answer

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

100%

Peisalla lamacron
Nar. ,202
Date
Name
Section
galauvinstenos
Pre-Laboratory Assignment
1. Read an authoritative source for a discussion of preparing and interpreting a graph, especially the
extrapolation of a curve.
2. A student determined the calorimeter constant of the calorimeter, using the procedure described in
this module. The student added 50.00 mL of cold water to 50.00 mL of heated, distilled water in a
Styrofoam cup. The initial temperature of the cold water was 21.00 °C and of the hot water, 29.15 °C.
The maximum temperature of the mixture was found to be 24.81 °C. Assume the density of water is
1.00 g mL¯1 and the specific heat is 4.184 J g¯K
gniwollot bns of toh bine
botollon ot ssb om
(1) Determine the AT for the hot water and the cold water.
mit
At cold water = 24.8I- 21,00 = 3.81°COn
आनी
AT hot = 24.81- 29.15 =
- 4.34°C
60.ES
2.0
0LES
8.
0.1
0.0
If
0.8
28.5S
Calculate N
0.4
08.TT
gabxim
02
4.34°C
hot water
81
answer
cold water
381°C
dogesn boximnu sdt to anuiSTeqmad mom odt ontmoiab bns sisb ami
answer
(2) Calculate the heat lost by the hot water.
N-C
4.34°C
answer
lgsg slt mot TA (S)
(3) Calculate the heat gained by the cold water.
3.810C
answer
Custom page
1988 Cengage Learning

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