Hot water mass: m hot = V 1"p Total water mass: m total = V zp * 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 = (q hot + q meli) fsion Heat absorbed per mole: * H fas molesmelt % difference from accepted value:

Chemistry In Focus
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Chapter9: Energy For Today
Section: Chapter Questions
Problem 51E: The second law of thermodynamics has been called the arrow of time. Explain why this is so.
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From the given, calculate:

1. Heat gained by melt water

2. ∆Hfus 

(show all calculations)

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)
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)
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:
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:
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