Question

Question 14 of 39 > How much heat is required to raise the temperature of 26.2 g of water from -35 "C to 155 "C? heat: about Careers vacy policy 11:24 SAMSUNG

Answer 1

Answers:

Calculations are given in image

14. q= 4.975 kcal

15. delT= 66.086 oC

16. m = 28.8 g

18. p = 0.75 atm

22. V= 0.270 L

In spertific Heat capacita q Mgo = Hisa i llegie. c = Heat capacity for specific heat capauty). G = Heat required. AT = change in temperature Given; T = -35°. Te = 155°C). AT = 19000) q = 4182 9 kgt oft & 1904 9.7945807 1557 for 26.29% Hoi q= 794580 ] 149 x 0.0262 kg 1 9 = 20817.9967 I cal = 4.1847 a = 20817 996 = 4975°6 cal. a . = by 1000 4.184 9=4.975 kaal

15) ez o.lled 9,9 004 ; 9: 205 cal for 28.2 g g iron ot x 2 0 c=o.ll cal gh of" x 28.2 g C = 3.109 cal och reggo 205 col (TA) [ 3-102 ekal get 66.086 cl ST-

(16) liven; S: 23.80 q = sluż 2597 m=2; c= 0.385.] bagay Heat capacity q ne copper. & , a n copper. CE MCAT) ! 200 E s cº1259) 0-385 7 Agt oct = 259] jo79.mx 23.3°C m= 2590 23.304 x 0.385 t meg tot 259 8.9705 223. 0 0 m= 28.8 9

Hig Boyle's Law. | Pic, a Pele =latm P2 - 13.30h. = 3.6L. bio sa V, :? : The Rivi P2 est = 1 an 4 3.6L 13.3 a tota = HQ = 3.6 L 13-3 1 = 0.270 L2 latm = 760 mmHg. 5.70 x 100 = 570 mm Hg. = 570 760 - 0.75 [p = 0.75 atm