Question

Datasheet Part 2 1. Mass of empty cups 6.48 2. Mass of cups plus 100 mL warm water 116.84 3. Initial temperature of warm water 366.0 7. Final temperature of water plus melted ice 5. Mass of water in calorimeter 18.9 21.2.139.05 29.27 145.53 - Mass of ice added Total mass Cups + ait H2O in the end Calculations - Part 2 For this system, note that AH calorimeter + 4Hwarm water + AHice water + AHfusion = 0 For the warm water and ice water, AH = m CAT. You may take c = 4.18 J/gºC for both. Assume AH calorimeter is negligible. 1. In the first part of the calculations, be sure to use the correct mass or water (or melted ice): a. Using your data, calculate Awarm water SIP

Answer 1

1. Using the mass of water (mass of cup+ water minus mass of cup ) , mass of ice and the temperature difference

$\Delta$H _{warm water} = -7.88 KJ and $\Delta$ H _{ice water} = - 2.09 KJ

Since the $\Delta$ H _calorimeter was negligible so $\Delta$ H _fusion was calculated using above results as 9.98 KJ

2. $\Delta$ H _fusionper gram of ice = 340.98 J/ g or 0.34 KJ /g

3.    $\Delta$ H _fusionper mole = 6.14 KJ / mole

4 . By all this the percentage difference was 2.16 %

(= 4.18 J1gc 0 0 mo ar a) for warm water mass of compty cup = 6.48 g mass of up + 100ml warm water = 116.84 g mass of 100 ml warm water = 116.34 - 6:48 m = 110.36 G AT = final komp - Duilial temp. - 18.9°C - 360° C -17.1°C Al warm water = MCAT = 110.36g X 4.18 I lg c X (-17. 1° C) -7888.31 J -7.88 kJ b) for ice water mass of see = 29279 IT = - 17.1°C Mice water = mc AT = 29.27 g xu. 18 319° 0 X (-17.1°c) = -2,092.16 I = -2.09 kJ C) Видеабил Malowmeter is megligible and In culorimeter & Alwarm water & Alike winter & Duferino

Continue C) Onwalm water & Skike water t All fusion o Alfason =- Hwarm water - I like water -- |-7888.31 J) - (-2092.165) - 9980.47] or 9.98 kJ 2. Difusion / gram of ice = 9980.47 I 29. 27 g = 340.985/9. 3. Alfason per nde =D Heusion X molar mass of ice gramgice - 340.98 Jlg x 18.02 g/mol - 6luu. U6 Il mol ol 6.14 kJ/mole

value - Absolute value x 100 I 4 % difference - experimental value - Absolute value Absolute value Absolute or accepted value = 6.01 kJ /mol experimental value = 6.14 kJ/mol % difference = 6.14 - 6. 0 x100 6.01 = 2.16%