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 122CP: As part of a science project, you study traffic patterns in your city at an intersection in the...
Related questions
Question
From the given, calculate:
1. Melt water temperature change
2. Heat lost by water
(show all calculations)
![Hot water mass: m hot = V1'p
%3D
4
Total water mass: m total = V2°p
Melt water mass: m melt = m total – m hot
Hot water temperature change: AT hot = T2– T1
Melt water temperature change: AT melt = T2 – 0.0 °C
Heat lost by hot water: q hot = Cp · m hot · AT hot
Heat gained by melt water: q melt = Cp · m melt · AT melt
Heat absorbed by melting ice: q fusion = (g hot + q melt)
9.
fusion
Heat absorbed per mole: ² H fus
molesmelt
% difference from accepted value:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47374bbc-3653-4105-ac48-6b4ae8c84a78%2F286dd2ff-1127-48ac-995c-50191b7b4bcd%2Fllzsf7_processed.png&w=3840&q=75)
Transcribed Image Text:Hot water mass: m hot = V1'p
%3D
4
Total water mass: m total = V2°p
Melt water mass: m melt = m total – m hot
Hot water temperature change: AT hot = T2– T1
Melt water temperature change: AT melt = T2 – 0.0 °C
Heat lost by hot water: q hot = Cp · m hot · AT hot
Heat gained by melt water: q melt = Cp · m melt · AT melt
Heat absorbed by melting ice: q fusion = (g hot + q melt)
9.
fusion
Heat absorbed per mole: ² H fus
molesmelt
% difference from accepted value:
![Time (s)
T (°C)
Time (s)
T (°C)
Time (s)
T (°C)
40°C
180
20.1°C
360
4.6°C
20
39.6°C
200
17.5°C
380
4.6°C
40
39.2°C
220
13.9°C
400
4°C
60
37°C
240
10°C
420
4°C
80
33.7°C
260
9.8°C
440
4°C
100
31.4°C
280
9.1°C
460
4°C
120
29.6°C
300
8.3°C
480
4°C
140
25.3°C
320
5°C
500
4°C
160
22°C
340
4.8°C
520
4°C
Hot water volume, V,
Final water volume, V2
125 mL
100 mL
Hot water initial
40 °C
Final water
18 °C
temperature, T,
temperature, T2
p40(H20) = 0.9932 g/mL
T.V. = 6.01 kJ/mol
Cp of water = 4.184 J/g°C
% difference from accepted value (% error) = T.V. - E.V. x 100
Moles melt = m melt /MMH,0
T.V.
Heat absorbed per mole = q fusion/ moles melt (E.V.) =
kI/mole (1k) =100OJ)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47374bbc-3653-4105-ac48-6b4ae8c84a78%2F286dd2ff-1127-48ac-995c-50191b7b4bcd%2Fjvo9h7i_processed.png&w=3840&q=75)
Transcribed Image Text:Time (s)
T (°C)
Time (s)
T (°C)
Time (s)
T (°C)
40°C
180
20.1°C
360
4.6°C
20
39.6°C
200
17.5°C
380
4.6°C
40
39.2°C
220
13.9°C
400
4°C
60
37°C
240
10°C
420
4°C
80
33.7°C
260
9.8°C
440
4°C
100
31.4°C
280
9.1°C
460
4°C
120
29.6°C
300
8.3°C
480
4°C
140
25.3°C
320
5°C
500
4°C
160
22°C
340
4.8°C
520
4°C
Hot water volume, V,
Final water volume, V2
125 mL
100 mL
Hot water initial
40 °C
Final water
18 °C
temperature, T,
temperature, T2
p40(H20) = 0.9932 g/mL
T.V. = 6.01 kJ/mol
Cp of water = 4.184 J/g°C
% difference from accepted value (% error) = T.V. - E.V. x 100
Moles melt = m melt /MMH,0
T.V.
Heat absorbed per mole = q fusion/ moles melt (E.V.) =
kI/mole (1k) =100OJ)
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