± Enthalpy of a Phase Change Part A Calculate the enthalpy change, 11. for the p

Question

± Enthalpy of a Phase Change Part A Calculate the enthalpy change, 11. for the process in which 147 g of water is converted from Iquid at 1.5°Ctovapor at 25.0°C For water, 1, 44.0 kJ/mol at 25.0 °C and C.-4.18 J/(g·°C) for H20(1). Express your answer to three significant figures and include the appropriate units Heat, q, is energy transferred between a system and its change where C, is specific heat and m is mass Hints Heat can also be transferred at a constant temperature when there is a change in state. For a process that involves a phase change 17: 1 Value Units where, n is the number of moles and 11 isthe enthalpy of fusion, vaporization, or sublimation The following table provides the specific heat and Part B of ioe at -19.2 C can be completely converted to liquid at 24.0 C if the available heat for this process is 5.4710 4.18 44.0 or ice, use a specific heat of 2.01 J/(g·°C)sd H-= 601kJ/mol Express your answer to three si 2.01 6.01 Hints Value Units My Answers Glive Up Con F2

Explanation / Answer

A)

m = 14.7 g of water coverted fom liquid 1.5°C to vapor at 25°C

Qliquid = m*C*(Tf-Ti) = 14.7*4.18*(25-1.5) = 1443.981 J

Qvapor = m*LH = 14.7/18*44*10^3 = 35933.33 J

Qtotal = 1443.981 +35933.33

Qtotal = 37377.311 J = 37.38 kJ

B)

Qrequirement

Qice = m*Cice * (0-Ti) = m*2.01*(0--19.2) =38.59 * m

Qmelting = m*LH = m/18 * 6.01*10^3 = 333.88*m

Qwater = m*Cwater*(Tf-0) = m*4.18*(24-0) = 100.32m

Qtotal = 38.59 * m+ 333.88*m+ 100.32m

Qtotal = (38.59+333.88+100.32) m

Q = 5.47*10^3 kJ = 5.47*10^6

5.47*10^6 = (38.59+333.88+100.32) m

m = (5.47*10^6) / ( (38.59+333.88+100.32))

m = 11569.61 g = 11.57 kg

sig fig 3 --> 11.6 kg

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