Question

if 0.15 kg of ice at -10 Celcius is placed in .6 kg of water at 50 Celcius, what is the final temperature? Assume the system is isolated.

Answer 1

Neat absorbtion by ice (ice) → Cice water) - → Water -10°C . 0° 1 0°C o final temperature (IP) Q = MC; AT Q = mL Q₂ = MCW ATZ Heat lost boy water at 50°c - Q₂ = MCW AT m = 0.15 kg , M = 0.6kg, L= 336 Kolkg , C; = 2.11g/kg-c Cw= 4.2 kJ/kg-°C AT, = 0=(-10%) = 10°c, DT2 = Tf - 0° = Tf BT3 = (50°C-T) Heat lost = Heat gained MCw (50°C-Tf) = mę (10°c) +mL +mCw TE = (M+ m) C Tq = MCW (50 c) = m C; (100c) & mL Tf = 0.6x4.2 x 50 – 0.15x2.1X10 - 0.15* 33 6 (0.6 +0.15) 4.2 T5 = 23°C]