2.50 Kg block of copper at 177 o c is brought into contact with a second 5.00 Kg

Question

2.50 Kg block of copper at 177oc is brought into contact with a second 5.00 Kg block of copper at 27oc. assume that there is no loss of heat. a-what is the final temperature of the two blocks? b- find the total entropy change of the two block system ? 2.50 Kg block of copper at 177oc is brought into contact with a second 5.00 Kg block of copper at 27oc. assume that there is no loss of heat. a-what is the final temperature of the two blocks? b- find the total entropy change of the two block system ?

Explanation / Answer

m1 = 2.50 Kg = 2.50*10^3 gm m2 = 5 Kg = 5*10^3 gm t1 = 177 degreeC = 450 K t2 = 27 degreeC = 300 K Specific heat of copper S = 0.386 J/gm K (a) Le t be the final temperature of the blocks. From law of calorimetry                       m1S(t1 - t) = m2S(t - t2)                           2.50(177 - t) = 5(t - 27)                                     7.5 t = 577.5                                          t = 77 degreeC = 350 K (b) Total change in entropy = m1S ln(350/450) + m2S(350/300)                                      = - 242.51 + 297.51 = 55 J/K

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