Question

The following information is given for water at 1 atm: boiling point = 100.00 °C melting point = 0.000 °C specific heat gas = 2.010 J/gºC specific heat liquid = 4.184 J/gºC AHvap (100.00 °C) = 2.259x103 J/g AHfus (0.000 °C) = 333.5 J/g kJ of energy are needed to raise the A 26.80 g sample of liquid water is initially at 32.30 °C. If the sample is heated at constant pressure (P= 1 atm), temperature of the sample to 113.20 °C.

Answer 1

1. 
   Heat energy needed = Heat required to leaise temperature from 32.3°c to loc & Heat required to vapowise & Heat required for raise temperature from loc to 113.20 Heat energy ree@ed = (26-8)(4.184) (100-3223)+2.259x16x2.8 +26-8)(2.010) (11302-10) = 7591.3 + 6054):2+ 711 = 68843.5 Joule :: Energy nel@ed - 68.84 KJ & Specific heat. & in I gmc Note: Heat required to raise temperature = msos Heat needed for phase change = m the o change - mass ingm mit Heat Capacity massingm. in II gm