Question

How many grams of ice at -24.9 degrees Celsius can be completely converted to liquid at 5.1 degrees Celsius if the available heat for this process is 4.22x10^3 kJ?

For ice, use a specific heat of 2.01J/(gxC) and enthalpy of fusion of 6.01 kJ/mol.

Answer 1

Answert Given that, Initial temperatude --24.9°C. Final temperatude - 5,1°C. ? = 4022x103KJ. Specific heat -2.01.5/99 off: 601 kJ/mol. Here, Ice (-24.9°c)-> ICP(0°C) water (oc)—watex (5.10) ในเ: hm 567 9, = dg x2-015/4° (0-624.9°c) 4= 50.04 X J. q2 = notte. 2 (18-0169mod boole 22 = 3348J for water 93= mSoT (5=401845/9c) 13 = xgx 40184 5/'c(501-60) 3=213x را هم q = 9, +q2tq

.6.22X103 KJ = 50.04&J + 3 3430+21.3x 4122X100 RT = $450.04 +334+ 21.3) 7 4:22 x103 K9 = 405.34 x J * = 4.22 X10° KJ . 405.345 x=1.041x104 g These fode, mass of ice - X