embry PHYS102 module 5 discuussion latest 2016 march

| April 14, 2018

Example: Calculate the amount of heat required to raise 0.550 kg of ice at -20oC to water at 20oC.Ice at -20 oC to ice at 0 oC: cice = 2090 J/(kg oC) Q1 = m cice∆T = (.55 kg)(2090 J/(kg oC)(0 – (-20)) = 22990 JIce at 0 oC to water at 0 oC: Lf =335,000 J/kgQ2 = m Lf = (.55 kg)(335000 J/kg) = 184250 J Note: This change occurs with no change in temperature.Water at 0 oC to water at 20 oC: cwater = 4186 J/(kg oC) Q3 = m cwater∆T = (.55 kg)(4186 J/(kg oC))(20 – 0) = 46046 JQTotal = Q1 + Q2 + Q3 = (22990 J) + (184250 J) + (46046 J) = 253286 J = 253000 J

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