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Law of Heat Exchange

Law of Heat Exchange. “governs all thermal interactions”. Thermal Energy (Q) always flow from the higher temp. object to the lower temp. object. The higher temp. object loses TE (-Q) and the lower temp. object gains TE (+Q).

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Law of Heat Exchange

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  1. Law of Heat Exchange “governs all thermal interactions” • Thermal Energy (Q) always flow from the higher • temp. object to the lower temp. object. The higher • temp. object loses TE (-Q) and the lower temp. object • gains TE (+Q). • The TE lost by the higher temp. object (QL) equals the • TE gained by the lower temp. object (Qg) . QL + Qg = 0

  2. Q = mcDt remember, Dt = tf - ti QL + Qg = 0 so, mcDtL + mcDtg = 0 or, mc(tf – ti)L + mc(tf – ti)g = 0

  3. mc = .500 kg mw = 1.00kg QL = -2000 J Qg = +2000 J Dtc = ? Dtw = ? 500. g of hot copper are placed in 1.00 L of cool water and the copper loses 2000 J of TE. What is the change in temperature of the copper and water. 1 g = 1 ml = 1 cm3 QL = mcDtc Dt = QL/mcc Dt = -2000 J/(.50 kgx 385 J/kg C) Dt = -10.4o C Qg = mcDtw Dt = Qg/mcw Dt = +2000 J/(1.00 kg x 4180 J/kg C) Dt = +.478o C

  4. What is the final temp. of the mixture if the 500 g of copper are at 100o C and the liter of water are at 20o C? QL + Qg = 0 mcDtc + mcDtw = 0 expand “Q” mc(tf – ti)c + mc(tf – ti)w = 0 expand Dt mctfc – mctic + mctfw – mctiw = 0 distribute (tf – ti) mctfc + mctfw = mctic + mctiw get tf and ti alone tf(mcc + mcw) = mctic + mctiw tfc = tfw so factor out tf = mctic + mctiw mcc + mcw Isolate for “tf”

  5. tf = .5kg(385J/kg C)100oC + 1kg(4180J/kg C)20oC .5kg(385J/kg C) + 1kg(4180J/kg C) Tf = 23.5o C

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