CHM110H5 Study Guide - Midterm Guide: Joule
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C12h22o11 per hour per hectare: 3. 295 x 105 kj h ha: a watt is a unit of power equal to one joule per second. 10,000 kj s ha x 3600 s h. Therefore, 3. 6 x107 kj h ha is the amount of solar energy: compare the amount of energy needed to produce of c12h22o11 (2) and the amount of solar energy to calculate the efficiency of photosynthesis (3): energy required percent efficiency= produce sucrose solar energy. = 3600kj h m: to calculate the square meters needed, compare the amount of kj per hour (3) and the amount of kj per hour per square meter (4): sqare meters needed= In this study, the minimum amount of surface area for a solar panel is determined to provide electricity for one house. On average household uses about 40 kwh of electricity per day. 1kw/m2 is equal to 1kj/sm2. So in one hour, 1kj/sm2 * 3600s will be equal to.