Question

Suppose this energy (984.1kJ) was deposited into a pitcher of ice containing 2100. g of water at 0 degree celsius and 1400. g of ice at 0 degree celsius. To what temperature would the resulting 3500. g of water increase following the absorption of the 984.1kJ?

Answer 1

Amount of heat absorted by the pitcher = gou. I KJ - 9.841 x 105 J Let the final temp. of Pitcher - Toc Amount of water = 2100 g specific heat of water (Ep, wr) = 4.184 Jlgoc AT= T-0°C = Toc a, 2100 x 4.104 T - 6 765.4 T I theat absorbed by 1400 g ice 1ч оо ханаuѕ + mx Clux GT = 1400 X 333.55 + 1450 x 4.194XT = 46690 + 1404.194T Total heat absorted 73G4 T +48 697 +1404.194T = 10190.504 T +48 6970 - 9 guio lolgo & 504 T- gouloo - 465970 loigo.584T = 517130 T- 517130 5. 74586 lolgo.504 = 50.75°C Final temp. = 50.75°C