From the given, calculate:1. Melt water mass2. Hot water temperature change Data SheetTime (s)T (°C)Time (s)T (°C)Time (s)T (°C)40°C18016.4°C3604.8°C2039.9°C39.6°C20012.2°C3804.5°C4.2°C4022010.4°C400608037°C33.6°C4204402407.9°C7.6°C4°C4°C26010030.3°C2805.8°C4604°C4°C5.5°C5°C4.9°C12026.1°C30048014016022.4°C3205005204°C18°C3404°CHot water volume,V1Hot water initial temperature, T1Final water volume,V2Final water temperature, T2100mL155.6 mL40°C4°CGiven: 25g ice cubes100g water at 40 °C (p40 = 0.9932 g/mL). Hot water mass: m hot = V1'p%3D4Total water mass: m total = V2°pMelt water mass: m melt = m total – m hotHot water temperature change: AT hot = T2– T1Melt water temperature change: AT melt = T2 – 0.0 °CHeat lost by hot water: q hot = Cp · m hot · AT hotHeat gained by melt water: q melt = Cp · m melt · AT meltHeat absorbed by melting ice: q fusion = (g hot + q melt)9.fusionHeat absorbed per mole: ² H fusmolesmelt% difference from accepted value:

Given: V1 is 100 mL. T1 is 40° C. V2 is 115.6 mL. T2 is 4° C. ρ(40° C) is 0.9932 g/mL. ρ(4° C) is 1…

Answered: From the given, calculate: 1. Melt…

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From the given, calculate: 1. Melt water mass 2. Hot water temperature change

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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Question

From the given, calculate:

1. Melt water mass

2. Hot water temperature change

Data Sheet
Time (s)
T (°C)
Time (s)
T (°C)
Time (s)
T (°C)
40°C
180
16.4°C
360
4.8°C
20
39.9°C
39.6°C
200
12.2°C
380
4.5°C
4.2°C
40
220
10.4°C
400
60
80
37°C
33.6°C
420
440
240
7.9°C
7.6°C
4°C
4°C
260
100
30.3°C
280
5.8°C
460
4°C
4°C
5.5°C
5°C
4.9°C
120
26.1°C
300
480
140
160
22.4°C
320
500
520
4°C
18°C
340
4°C
Hot water volume,V1
Hot water initial temperature, T1
Final water volume,V2
Final water temperature, T2
100mL
155.6 mL
40°C
4°C
Given: 25g ice cubes
100g water at 40 °C (p40 = 0.9932 g/mL).

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:

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