general chemistry 1

Question

At 1 atm, how much energy is required to heat 57.0 g H2O(s) at −18.0 ∘C to H2O(g) at 161.0 ∘C? Use the heat transfer constants found in this table.

answer: q= _____kJ

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 H 2 O (ice)	2.087 J/(g·°C) *	37.60 J/(mol·°C) *
Specific heat of liquid H 2 O (water)	4.184 J/(g·°C) *	75.37 J/(mol·°C) *
Specific heat of gaseous H 2 O (steam)	2.000 J/(g·°C) *	36.03 J/(mol·°C) *

Answer 1

This is a step-wise problem. Each step calculates the energy (heat) needed for the particular change in question. The equations to be used are

q = mC∆T and q = m∆H

q = heat; m = mass; C = specific heat; ∆T = change in temperature and ∆H = heat for the phase change, either fusion or vaporization (see below).

Step 1: raise temp of solid water from -18º to zero degrees

q = mC∆T = (57.0 g)(2.087 J/gº)(18º) = 2141 J

Step 2: melt solid water at zero degrees - phase change and no change in temperature

q = m∆Hfusion = (57 g )(333.6 J/g) = 19,015 J

Step 3: raise temp of liquid water from zero degrees to 100 degrees

q = mC∆T = (57 g)(4.184 J/gº)(100º) = 23,849 J

Step 4: turn liquid water into steam (gas) at 100º - phase change and no change in temperature

q = m∆Hvap = (57 g)(2257 J/g) = 128,649 J

Step 5: raise the temperature of gaseous water from 100º to 161º

q = mC∆T = (57 g)(2.00 J/gº)(61º) = 6954 J

Sum all the heats to get...

2141 + 19015 + 23849 + 128649 + 6954 = 180,608 J = 181 kJ (3 sig. figs.)