CHM110H5 Study Guide - Quiz Guide: Kilowatt Hour, Joule

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13 Feb 2014
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Assignment 5: solar energy = 1 kw m2. 58. 42 mol x kj x = 329496. 99 kj of sunlight used to produce 20 kg c12h22o11. 20 kg/h/ha = 329496. 99 kj/(60 x 60)s/10000 m2. = 0. 00915 kw/m2 (since 1 kj/s/m2 = 1 kw/m2) 2 x 100% = 0. 915% of sunlight used to produce the sucrose. 1. 0 kw/m2: 15% efficient in converting sunlight to electricity. 40 kwh x kj x = 144000 kj of electricity energy needed per day. 144000 kj = 0. 15 x x (where x represents the solar energy) x = 960000 kj of solar energy. We also know that i kj/s/m2 = 1 kw/m2. Surface area = __number of kj/h__ number of kj/h/m2. = 120000 kj/h = 33. 33 m2 of the surface area of the solar panels. 3600 kj/h/m2: we know from the problem , that the solar panels are 15% efficient in converting sunlight to electricity and 40 kwh of electricity is used per day.

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