Consider heating solid water (ice) until it becomes liquid and then gas (steam).
ID: 867363 • Letter: C
Question
Consider heating solid water (ice) until it becomes liquid and then gas (steam). (Figure 1) Alternatively, consider the reverse process, cooling steam until it becomes water and, finally, ice. (Figure 2) In each case, two types of transitions occur, those involving a temperature change with no change in phase (shown by the diagonal line segments on the graphs) and those at constant temperature with a change in phase (shown by horizontal line segments on the graphs). Equations The heat energy associated with a change in temperature that does not involve a change in phase is given by q = ms delta T where q is heat in joules, m is mass in grams, s is specific heat in joules per gram-degree Celsius, J/(g degree C), and delta T is the temperature change in degrees Celsius. The heat energy associated with a change in phase at constant temperature is given by q = mdelta H where q is heat in joules, m is mass in grams, and delta H is the enthalpy in joules per gram. Physical constants. The constants for H2O are shown here: Specific heat of ice: SiCe = 2.09 J/(g.d egree C) Specific heat of liquid water: s water = 4.18 J/(g .degree C) Enthalpy of fusion (H2 0(s)rightarrow H2o(1)): delta Hfus = 334 J/g Enthalpy of vaporization (H2 O(1) rightarrow H2 O(g)): delta Hvap = 2250 J/g How much heat energy, in kilojoules, is required to convert 76.0g of ice at -18.0 degree C to water at 25.0 degree C ? Express your answer to three significant figures and include the appropriate units.Explanation / Answer
Q = heat change for conversion of ice at -18 oC to ice at 0 oC + heat change for conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 25 oC
Amount of heat released , Q = mcdt + mL + mc'dt
= m(cdt + L + c'dt' )
Where
m = mass of water = 76.0 g
c' = Specific heat of water = 4.18 J/g degree C
c = Specific heat of ice= 2.09 J/g degree C
L'= Heat of fusion of ice OR enthalpy of fusion = 334J/g
dt' = 25 -0 =25 oC
dt = 0-(-18)=18 oC
Plug the values we get Q = m(cdt + L + c'dt ')
= 76.0 x [ (2.09 x 18) + 334 + ( 4.18 x 25 )]
= 76.0 x 476.12
= 36185 J
= 36.2 KJ
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