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22 June, 03:05

piston cylinder has 5kg water at 500kPa and 300c. Heated by 100V and 4A that operates for 500 secs. It's cooled to a saturated liquid. what is the boundary work and heat lost?

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  1. 22 June, 06:09
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    Answer: boundary work is 1.33kJ and heat lost is 200kJ

    Explanation: mass of water m = 5kg

    Molar mass of water mm = 18kg/mol

    No of moles n = 5/18 = 0.28

    Pressure P = 500x10^3

    Temperature T = 300°c = 573K

    R = 8.314pa. m3/mol/k a constant.

    Using PV = nRT

    V = nRT/P

    V = (0.28x8.314x573) : (500x10^3)

    V = 0.00266m3

    Work on boundary w = PV

    W = 500x10^3 x 0.00266 = 1330 J

    = 1.33kJ

    Heat lost on cooling is equal to electrical heat Q used to heat the water initially

    Q = Ivt

    I = current = 4A

    v = voltage = 100

    T = time = 500sec

    Q = 4x100x500 = 200000 = 200kJ
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