CU PHYS221 Practice Test 3 Solution - vC

Find q know m 045 kg c water 4186 jkg ºc t i 220 ºc

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Find: Q Know: m = 0.45 kg c water = 4,186 J/kg-ºC T i = 22.0 ºC T boiling = 100.0 ºC L v-water = +2.26 x 10 6 J/kg c steam = 2,010 J/kg-ºC T f = 160.0 ºC Sketch : Equation(s) : Q = mcΔT Q = mL Solve: This is a specific heat problem involving a phase change NOTE : The mass of water = mass of steam if there are no losses The problem has to be done in three parts: (a) Raise temperature of 0.45 kg water from 22.0 ºC to 100.0 ºC. Use the equation “Q = mcΔT” where the change is temperature “ΔT” is: ΔT = (T f − T i ) = (100.0°C) – (22.0°C) = 78.0 ºC 7 Version C 100 ºC Ice Ice & Water T Water Water and Steam Steam 0 ºC T initial = 22.0ºC T final = 160.0ºC
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PHYS221 Practice Test 3 To find the heat for this part of the problem, use the equation “Q = mcΔT” with the known values of “m water = 0.45 kg” “c water = 4186 J/kg-ºC” and “ΔT = 78.0 ºC” Q = m ice c ice ΔT = (0.45 kg)(4186 J/kg-ºC)(78.0 ºC) = 1.47 x 10 5 J (b) Phase change 0.45 kg water to 0.45 kg steam Use the equation “Q = mL f ” with the known values “m water = 0.45 kg” and L v-water = +2.26 x 10 6 J/kg Q = m water L v-water = (0.45 kg)(2.26 x 10 6 J/kg) = 1.02 x 10 6 J NOTE: The “L v-water ” is positive (+) because we are adding heat to the water! (c) Raise temperature of 0.45 kg steam from 100.0 ºC to 160.0 ºC Use the equation “Q = mcΔT” where the change is temperature “ΔT” is: ΔT = (T f − T i ) = (160.0°C) – (100.0 ºC) = 60.0 ºC To find the heat for this part of the problem, use the equation “Q = mcΔT” with the known values of “m steam = 0.45 kg” “c steam = 2010 J/kg-ºC” and “ΔT = 60.0 ºC” Q = m steam c steam ΔT = (0.45 kg)(2010 J/kg-ºC)(60.0 ºC) = 5.43 x 10 4 J The total heat added is sum of the heat required in parts (a), (b) and (c): Q total = (1.47 x 10 5 J) + (1.02 x 10 6 J) + (5.43 x 10 4 J) = +1.22 x 10 6 J
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