Recent content by portofino

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    Thermodynamics - efficiency from pv diagram

    i think it is a ratio of e_carnot/e where e = 17% ^^^^that ratio is correct, e_carnot/e = 83.33/17 = 4.90
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    Thermodynamics - efficiency from pv diagram

    let 6atm, 2L = 1, 6atm,6L = 2, 3atm, 6L = 3, 3atm, 2L = 4: 1-->2 isobaric expansion Q = nC_pT = n(5R/2)(PV/nR) = 5/2(PV) = 5/2(607950)(0.006-0.002) = 6079.5 2-->3 isovolumetric compression Q = n(3R/2)(PV/nR) = 3/2(PV) = 3/2(303975-607950)(0.006) = -2735.8 3-->4 isbaric compression Q =...
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    Thermodynamics - efficiency from pv diagram

    i still can't seem to get it, how do i find the q for the isvolumetric paths when Q = 3/2(P*deltaV) where pressure changes over each pf the two isovolumetric paths? help please
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    Thermodynamics - efficiency from pv diagram

    i tried what you said and got Q = 6079.5 joules for top isobaric, and Q = -1823.85 joules for bottom isobaric so e = W/Q_h = 1215/(6079.5-1823.85) = 1215/4255.65 = 28.5% which was wrong, what now?
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    Thermodynamics - engine efficiency

    oh i should've been more specific, i do know people who go to umbc, i am actually posting on behalf of one my old school mates who goes to umbc, he works nights and cant get stuff done btw 24.3 worked
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    Thermodynamics - engine efficiency

    nah i go to old dominion, i already tried W/Q_h where W = 350 joules, my original Q_h was wrong i used temp not Q, as for Net = 691.5-748.3+748.3-345.8 = 345.8, so should e = 350/345.8 = 101.2 % or 350/350 = 100%? or should it be Q_h = 691.5 + 748.3 = 1439.8 so e = 350/1439.8 = 24.3%?
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    Thermodynamics - engine efficiency

    using (Q_h - Q_c)/Q_h = (600 - 300)/600 = 50% was not correct? what am i doing wrong?
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    Thermodynamics - engine efficiency

    the question wants me to use a ratio of work and heat energy though, why should i use (Q_h - Q_c)/Q_h? so should i be getting 50% then is use the efficiency equation you suggest?
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    Thermodynamics - efficiency from pv diagram

    Homework Statement Determine the efficiency for the cycle shown in the figure, using the definition. Assume that the gas in the cycle is ideal monatomic gas. Compare with the efficiency of a Carnot engine operating between the same temperature extremes. see attachment Homework...
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    Thermodynamics - engine efficiency

    Homework Statement A reversible engine contains 0.20 mol of ideal monatomic gas, initially at 600 K and confined to 2.0 L. The gas undergoes the following cycle: Isothermal expansion to 4.0 L. Isovolumic cooling to 300 K. Isothermal compression to 2.0 L. Isovolumic heating to 600 K...
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    Thermodynamics - hot item placed in water

    sorry to double post, just wanted to let you all know that the above values i got were in fact correct. thanks for the help
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    Thermodynamics - hot item placed in water

    using the above equation i got this: Tf_w = Tf_al = 24.5 = 24.5 + 273 = 297.5 kelvin Ti_w = 19 = 19 + 273 = 292 kelvin find Ti_al [(12)(4184)(297.5-292)]_water + [(0.9)(900)(297.5 -Ti)]_aluminum = 0 271644 + 240975 - 810Ti = 0 810Ti = 517119 Ti = 638.41 kelvin, so for part one Ti in celsius =...
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    Thermodynamics - hot item placed in water

    is that the same case for the water, so instead of 5.5 it should also be T = (24.5 - Ti), what happens to the 19 degrees stated in the problem?
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    Thermodynamics - hot item placed in water

    i see what you're saying topher, so T should then be (24.5 - Ti)?
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