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H and SJ. f. 2) for the non-flow energy equation. Similady, work is given by sJ. 1 W2 = sJ. h. u, the non-flow energy equation. u = if q = w = 0; the law of conservation of energy. Corollary 3 The continuous output of work from a system with zero input of energy is impossible, since 1liw = if 1liq = 0; a perpetual motion machine of the first kind. 1. 1, determine the quantity of heat supplied when the interna! energy of the gas (a) remains constant, and (b) increases by 100 kJ. u. 2. A fluid contained in a piston-in-cylinder device receives 150 kJ of mechanica1 energy by means of a paddle whee1, together with 50 kJ by heat transfer.

This work could then be supplied to a second system which converts it completely and continuously back to heat, but at a higher temperature, and delivers it to a second reservoir (type C): such a work-heat conversion does not violate any laws of thermodynamics. The sole overall effect of these coupled systems (B + C) would be the transfer of heat from a low-temperature reservoir to a high-temperature reservoir, which violates the Clausius statement. By similar reasoning, the performance of a heat engine (type D) in which heat q2 can flow unaided from the sink to the source, in contradiction to the Clausius statement, can be seen to violate the Kelvin-Planck statement since the sink becomes redundant, thus the two statements are equivalent.

H. u, the non-flow energy equation. u = if q = w = 0; the law of conservation of energy. Corollary 3 The continuous output of work from a system with zero input of energy is impossible, since 1liw = if 1liq = 0; a perpetual motion machine of the first kind. 1. 1, determine the quantity of heat supplied when the interna! energy of the gas (a) remains constant, and (b) increases by 100 kJ. u. 2. A fluid contained in a piston-in-cylinder device receives 150 kJ of mechanica1 energy by means of a paddle whee1, together with 50 kJ by heat transfer.

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