(A) Suppose you cool a 16 oz can of beer by putting it in a cooler containing ic
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Question
(A) Suppose you cool a 16 oz can of beer by putting it in a cooler containing ice/water slush at 0 C. Use regular ounces for the weight (i.e. 16 oz = 1 pound) of the beer and NOT fluid ounces. Assume the beer can be treated as water and that the beer can contains 17 g of aluminum. How much heat is lost from the can of beer by the time it reaches 0 C if it started at room temperature of 22 C? The heat capacity of water is 4.18 J/g-K and the heat capacity of aluminum is 0.90 J/g-K. Assume that the cooler remains at 0 C and does not change temperature with the addition of the beer can. 43.15, 42.05, 41.71, 42.34, 42.93 (in kJ)
(B) Calculate the H of reaction for the metabolism of 38 g of glucose (C 6H 12O 6) with excess oxygen to form CO 2 and water if the H of formation for C 6H12O 6, H 2O (l) and CO 2 are -1260, -286 and -394 kJ/mol, respectively? Enter a numerical value below and be sure to include a minus sign if needed. The units are kJ
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Explanation / Answer
B) C12H22O11(aq) + 12O2(g) ® 12CO2(g) + 11H2O(l)
DHo = {12(-394) + 11(-286)} – {-38/342(molar mass of sugar)1260}
= -7851 + 140 = -7711 kJ mol-1 Thus, each 1.0 g liberates -5625/342 = 16 kJ of energy.
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