1:11 File Details CHEM 120 - Fund of Col Chem # 3765 - Spring 2022 = 65°C, final Tis 87°) The purpose of this experiment

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1 11 File Details Chem 120 Fund Of Col Chem 3765 Spring 2022 65 C Final Tis 87 The Purpose Of This Experiment 1 (36.97 KiB) Viewed 46 times

1:11 File Details CHEM 120 - Fund of Col Chem # 3765 - Spring 2022 = 65°C, final Tis 87°) The purpose of this experiment is to measure the number of calories needed to melt ice. This is known as the "heat of fusion" of ice. (heat of fusion) H₂O(ice) at 0°C H₂O(liquid) at 0°C Note - The melting process occurs at a constant temperature - there is no AT. It is surprising that during the phase change (solid to liquid), heat must be added yet the temperature does not increase. The added energy is not lost, it is used to overcome the intermolecular forces holding the solid together. The following experimental data will allow you to calculate the Heat of Fusion of Ice. 125 grams of warm, 54.6°C water, was poured into a temperature insulating cup. Next, 35.2 grams of ice at 0°C was sealed in a small plastic bag and placed in the cup. The plastic bag kept the warm water and ice separate from each other while the ice was melting. At the precise moment that the last bit of ice melted, the warm water had cooled to 31.2°C. Calculations - Heat of Fusion of Ice 1. How many calories of heat were lost by the 125 g of warm water in cooling from 54.6°C to 31.2°C? Given: Heat-> (grams of water) (1.00 calorie/g °CX AT) *The heat lost by the warm water heat used to melt the ice (answer: 2,925 calories = 2,930 cal.) (heat of fusion) 2. Use the calculated value from #1 (above) and the mass of ice to calculate the heat of fusion per gram. (answer: 83.2 calories/g) 3. If the accepted value for the heat of fusion of ice is 80.0 calories-per- das * Previous Next 13 Dashboard Notifications Calendar To Do Inbox 64