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physics mc(急)
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http://upload.lsforum.net/users/public/h80555m143.gif http://upload.lsforum.net/users/public/l169556e143.gif http://upload.lsforum.net/users/public/n549527l143.gif 煩請解釋。謝謝 更新: 感謝回答28和36。第18題,我都係計到呢個答案,不過ANS GIVEN係B。
最佳解答:
18. Use the one with a power of 88.9 w Latent heat of vapourization = 88.9x60/2.1 J/g = 2540 J/g = 2.54 x 10^6 J/kg The answer is option C The reason is that the heat loss to the surroundings would be negligible when a high power is used, but may contribute a significant portion to the heat supplied should a low power is used. Assumption: the heat loss to the surroundings is kept constant throughout the experiment. 28. Option B. Using PV/T = constant where P, V and T denote the pressure, volume and temperature of the gas. From X to Y, P is constant, hence, V and T are proportional. From Y to Z, V is constant, hence P and T are proportional From Z to X, when P decrease, V increase, T could be kept constant. 36. Use the ideal gas equation PV=nRT Let n1 and n2 be the initial no. of moles of the gas in containers X and Y respectively hence, n1 = 400V/R(273+200) = 400V/473R n2 = 100.(4V)/R(273+400) = 400V/673R Total mass of gas in moles = 400V/473R + 400V/673R After the tap is opened, let the common pressure be P If n1' and n2' be the new no. of moles in containers X and Y respectively then, n1' = PV/473R and n2' = P(4V)/673R Because n1 + n2 = n1' + n2' (conservation of mass) thus, 400V/473R + 400V/673R = PV/473R + P(4V)/673R i.e. 400/473 + 400/673 = P/473 + 4P)/673 solve for P gives P = 179 kPa The answer is option B 2012-07-21 16:31:53 補充： Sorry, I have misinterpreted Q18. The solution is For the 1st heater, 88.9 x 60 = 2.1L+Q For the 2nd heater, 42x60 = 0.9L + Q where L is the latent heat (in J/g) and Q is the heat loss to surroundings Subtracting the 2 equations: 46.9x60 = 1.2L L = 2.345 J/g = 2.35x10^6 J/kg 2012-07-21 16:33:10 補充： A typo in the last line: L = 2,345 J/g = 2.35 x 10^6 J/kg
physics mc(急)
發問:
http://upload.lsforum.net/users/public/h80555m143.gif http://upload.lsforum.net/users/public/l169556e143.gif http://upload.lsforum.net/users/public/n549527l143.gif 煩請解釋。謝謝 更新: 感謝回答28和36。第18題,我都係計到呢個答案,不過ANS GIVEN係B。
最佳解答:
18. Use the one with a power of 88.9 w Latent heat of vapourization = 88.9x60/2.1 J/g = 2540 J/g = 2.54 x 10^6 J/kg The answer is option C The reason is that the heat loss to the surroundings would be negligible when a high power is used, but may contribute a significant portion to the heat supplied should a low power is used. Assumption: the heat loss to the surroundings is kept constant throughout the experiment. 28. Option B. Using PV/T = constant where P, V and T denote the pressure, volume and temperature of the gas. From X to Y, P is constant, hence, V and T are proportional. From Y to Z, V is constant, hence P and T are proportional From Z to X, when P decrease, V increase, T could be kept constant. 36. Use the ideal gas equation PV=nRT Let n1 and n2 be the initial no. of moles of the gas in containers X and Y respectively hence, n1 = 400V/R(273+200) = 400V/473R n2 = 100.(4V)/R(273+400) = 400V/673R Total mass of gas in moles = 400V/473R + 400V/673R After the tap is opened, let the common pressure be P If n1' and n2' be the new no. of moles in containers X and Y respectively then, n1' = PV/473R and n2' = P(4V)/673R Because n1 + n2 = n1' + n2' (conservation of mass) thus, 400V/473R + 400V/673R = PV/473R + P(4V)/673R i.e. 400/473 + 400/673 = P/473 + 4P)/673 solve for P gives P = 179 kPa The answer is option B 2012-07-21 16:31:53 補充： Sorry, I have misinterpreted Q18. The solution is For the 1st heater, 88.9 x 60 = 2.1L+Q For the 2nd heater, 42x60 = 0.9L + Q where L is the latent heat (in J/g) and Q is the heat loss to surroundings Subtracting the 2 equations: 46.9x60 = 1.2L L = 2.345 J/g = 2.35x10^6 J/kg 2012-07-21 16:33:10 補充： A typo in the last line: L = 2,345 J/g = 2.35 x 10^6 J/kg
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