Consider heating solid water (ice) until it becomes liquid and then gas (steam).
ID: 1007527 • 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). 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, 5 is specific heat in joules per gram-degree Celsius. J/(g middot 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 = m delta H where q is heat in joules, m is mass in grams, and AH is the enthalpy in joules per gram. Physical constants The constants for H_2O are shown here: Specific heat of ice: s_water = 2.09 J/(g middot degree C) Specific heat of liquid water: s_water = 4.18 J/(g middot degree C) Enthalpy of fusion (H_2O(s) rightarrow H_2O(1)): delta H_fus = 334 J/g Enthalpy of vaporization (H_2O(l) rightarrow H_2O(g)): delta H_vap = 2250 J/g How much heat energy, in kilojoules, is required to convert 38.0 g 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 unit How long would it take for 1.50 mol of water at 100.0 degree C to be converted completely into steam if heat were added at a constant rate of 24.0 J/s? Express your answer to three significant figures and include the appropriate units.Explanation / Answer
In this case total three type of heat:
Q = [ mice ] [ ( Cvice ) ( 0.0 C - T1ice ) + ( H sub sf ) + ( Cvwater ( ( Twater2 - 0.0 C ) ]
Q= Q1+Q2+Q3
Q1= [ mice ] [ ( Cvice ) ( 0.0 C - T1ice )
= [38.0 g ][ ( 2.09 J/g - C ) ( 0.0 C - -18.0 C )
= 1429.56 J
Q2 =[ mice ]* ( H sub sf )
= 38.0 g* 333.7 J / g
=12692 J
Q3= =[ mice ]* (Cvwater ( ( Twater2 - 0.0 C ) ]
= 38.0 g*(4.18 J / g - C ) ( 25.0 C - 0.0 C ) ]
= 3971 J
Q = [1429.56 J+12692 J+3971 J]
=18092.56 J * 2250 j/g
= 60817.5 J
Given that;
24.0 J/ s
Time = total heat / 24.0 J/s
= 60817.5 J/24.0 J/S
=2534.0625 s
= 42.23 min
= 18.092 KJ
B.
Mass of water= 1.50 moles * 18.02 g/ moles
= 27.03 g
Q= mass of water * heat of vaporization
= 27.03 g*2250 J/ G
= 60795 J
Given that;
24 J/ s
Time = heat / 24.0 J/s
= 60795 J/24.0J/s
= 2533.125 S
= 42.22 min
=0.74 hr
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