An iceburg weighing 10 9 kg at a temperature of 0degrees C drifts into the Gulf

Question

An iceburg weighing 109kg at a temperature of 0degrees C drifts into the Gulf Stream, which is at a temperature of20 degrees C. After several weeks there remains only water at 20degrees C. What is the entropy change in the universe associatedwith the disappearance of the iceburg? The answer provided bythe book is 2.27 x 107 kcal/K. For the entropy change of the iceberg, I get 5.65 x109 kcal/K and for the water I get 5.45 x 109kcal/K. These figures do not work out when calculating the entropychange for the universe. Please help. Thanks. An iceburg weighing 109kg at a temperature of 0degrees C drifts into the Gulf Stream, which is at a temperature of20 degrees C. After several weeks there remains only water at 20degrees C. What is the entropy change in the universe associatedwith the disappearance of the iceburg? The answer provided bythe book is 2.27 x 107 kcal/K. For the entropy change of the iceberg, I get 5.65 x109 kcal/K and for the water I get 5.45 x 109kcal/K. These figures do not work out when calculating the entropychange for the universe. Please help. Thanks.

Explanation / Answer

First, heat lost by ocean = heat gained byiceberg = .           mL + mc T    =  109 * 79.7 + 109 * 1 * 20 = 99.7 x 109 kcal . For the iceberg... . delta S =   entropy for melting + entropy for warming to final temp = .              =   mL / T       +   m c ln ( T final / T initial) = .              = 109 * 79.7 / 273   +  109 * 1 * ln(293/273) = .               =   3.62642 x 108 kcal/K . For the ocean... . delta S   =   entropy due to heatlost to iceberg   = heat lost by ocean / T = .          =  - 99.7 x 109 kcal / 293 K = -3.40273 x108 kcal/K . Total for process = 3.62642 - 3.40273 =   0.2237 x 108  =   2.24 x107 kcal/K . My answer is slightly different than that of the book. It isprobably due to a bit of difference in rounding. You didnt say howyou calculated your answers, but check them against mine and seewhere the difference is.

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