Quations The heat energy associated with a change in temperature that does not i
ID: 790503 • Letter: Q
Question
Quations The heat energy associated with a change in temperature that does not involve a change in phase is given by where q is heat in joules, rn is mass in grams, s is specific heat in joules per grarn-degree Celsius,J/(g*C), and deltaT is the temperature change in degrees Celsius. The heat energy associated with a change in phase at constant temperature is given by q = rn*delta H where q is heat in joules, rn is mass in grams, and deltaH is the enthalpy in joules per gram. Physical constants: the constants for H20 are shown here: specific heat for ice = 2.09 J/(g*C) specific heat foe liquid water =4.18 J/(g*C) Enthalpy of fusion (H20(s) ->H20(I)): delta H fus = 334J/g Enthalpy of vaporization ( H20(l) >H20(g)): delta Hvap = 2250 J/g How much heat energy, in kilojoules, is required to convert 59.0mol of ice at 18.OC to water at 25.0 C ? Express your answer to three significant figures and include the appropriate units. long would it take for 1.50 mol of water at 100.0 C to be converted completely into stearn if heat were added at a constant rate of 17.0J/s ? Express your answer to three significant figures and include the appropriate units.Explanation / Answer
A)
How much heat energy, in kilojoules, is required to convert 59.0mol of ice at -18.0C to water at 25.0 C ?
Qice = m*Cice * (Tf-Ti)
Qlatent = m*LH
Qwater = m*Cwater *(Tf-Ti)
substitute
Qice = m*Cice * (Tf-Ti) = 59*18 * 2.01 * (0 - -18) = 38423.16J
Qlatent = m*LH = 59*18*334 = 354708
Qwater = m*Cwater *(Tf-Ti) = 59*18*4.184*(25-0) = 111085.2
Total Q = 38423.16 + 354708 + 111085.2
Q = 504216.36 J = 504.216 kJ
B) long would it take for 1.50 mol of water at 100.0 C to be converted completely into steam if heat were added at a constant rate of 17.0J/s ?
First, find Q
Q = m*LH = 1.5*18*2264.76 = 61148.52 J
rate is
P = Q/t
t = Q/P = 61148.52 / 17
t = 3596.971 seconds = 3596.971/3600 = almost 1 hour