Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.550 kg of water decreased from 109 °C to 38.5 °C.

Property	Value	Units
Melting point	0	°C
Boiling point	100.0	°C
ΔHfus	6.01	kJ/mol
ΔHvap	40.67	kJ/mol
cp (s)	37.1	J/mol · °C
cp (l)	75.3	J/mol · °C
cp (g)	33.6	J/mol · °C

Answer 1

The total energy changed can be determined as, 9 = 4, +92 +93 where, q = Heat required to convert 109°c gutom to 109°C Tapetano Steam vapour = m. Hy [ Hy lat Vapouri Zatim Hv [ THv = latent heat of [ m = mass of the - Water? G7557 4HY = 40.67 kJ/mes z 2 40.67 kJ/mos molecular I weight of 1 18 moito18 amel = 2.26 kJ/a m=0.550kg 20-550 x1600 g = 5500 -9, = 5508 x 2.26 K5/8 2loj = 1243 KJ. = 1243x10 J 12 = Energy required to convert steam 1. Water to the boiling point of water zm CpGAT [m=550g cp = specific heat of steam = 550 982.06 I tot = 37-1 I/mol'c. - 37.1 I 18 = 2.06 4 T = (109-100) g'c. (Page 1) = 10197 J. gic. Scanned with CamScanner

- tempe. 43= the energy required to lower rature from 180c to 38-5c. +5.3 Doc in - mcpeat I m 2 550 o a aple) = 75-3 J/mooc. 2 5509x 9:18.75€ x6154 = 141: 38 x103 J. 24.18 96. SIT =1180-385) 2 761.599] Total energy Å= 4, +9₂ +93 = $(1243 x103 +10197 +141-38 x103) 3 = 1.39467106 J = 1394.6 KJ. (page-2) Scanned with CamScanner