Question

How many grams of ice at -7.3°C can be completely converted to liquid at 10.5 °C if the available heat for this process is 5.37 * 10 ^ 3 * kJ? For ice, use a specific heat of 2.01 J/(g*^ C) and Delta H fus =6.01 kJ/mol

Answer 1

Let e gm ice can be e gmice at -7.3c converted, a xe gmice at ocy. re gm water at 1000cc ke gm water at occ 13 Cwatan = 4.18 cice = 2.ole 2fusion = Golka = 6010 I (Amol water = 18gm) - m O mol a Losion = 334 ? heat quailable = 5.37x103 kJ = 5.37x10by (1k9=103) heat = Hit H2 + H3 = (mCatlicet me t (mcoT) water S-374106 = (ex *****7.3) + ** 334 + (2x468x10g) a 2.0 a z = 13680 gm 13680 gm) (mass of ice = CScanned with CamScanner