Question

At 1 atm, how much energy is required to heat 73.0 g H2O(s)73.0 g H2O(s) at −14.0 ∘C−14.0 ∘C to H2O(g)H2O(g) at 159.0 ∘C?159.0 ∘C? Use the heat transfer constants found in this table.

Heat-transfer constants for H₂O at 1 atm

Quantity	per gram	per mole
Enthalpy of fusion	333.6 J/g	6010. J/mol
Enthalpy of vaporization	2257 J/g	40660 J/mol
Specific heat of solid H2O (ice)	2.087 J/(g·°C) *	37.60 J/(mol·°C) *
Specific heat of liquid H2O (water)	4.184 J/(g·°C) *	75.37 J/(mol·°C) *
Specific heat of gaseous H2O (steam)	2.000 J/(g·°C) *	36.03 J/(mol·°C) *

Answer 1

We have to take all steps tre H20 (3) od 10 (s) soz Hoole) SO3 H2O(1) du holg) ods. How -14°C 0°c o'c 10000 1000 15e total energy - AQ,+ AQ2 +oQg+oQy tods - mass (on) - 739 de enthalpy of fusion) = 333, 6 Jg SH2013) - 2.097 396 Jylenthalpy of Vaporisation SHO(1) = 4.184 Igle = 225750 SH₂O (8) = 2.000 Igo Here Civen le Q= mSuis) AT AT = 0c-(-14°c) = 73 * 2.087 5 g x 14°C - 14°C = 2132.9145 DQ2 = mit = 738 3 33.6 I g = 24352-8 J DQ3 = m Sole AT = 739 x 4.184 I J & * 100 73.9 4.1847 oo's SAT 1000-002 = 1000 l - 30543,2 J DQ4 = ml = 738 x 2257 JJV = 164761 T AQs = mSH2018) AT = 739 x 2.000gg c ' x 596 - 8614 J from total energy required = 2132.914 J +24352.83 +30543.2J +164701 J + 8614 J 230403.914 J E 230.4 KJ]