PLEASE ANSWER A-K !! 1. Solve the following problem: You will determine the Heat

Question

PLEASE ANSWER A-K !!
1. Solve the following problem: You will determine the Heat of Fusion of ice as follows. Weigh a dry calorimeter containing a magnetic stirring bar. Heat of Fusion of Ice Add the required amount of hot (60-70°C) water Time-Temperature Curve (Logger-Pro) Ice is being added and reweigh. Start stirring. Insert a temperature sensor and "continuously" record the temperature vs time. Initial temperature of hot water After about 30 seconds start adding pieces of ice until the temperature drops to between 20 and Final temperature 30°C Record this final temperature for at least 30 seconds. Make sure there is efficient stirring and all the ice melted. Finally weigh the calorimeter and final contents NOTE: the ice must be at 0°C so, if taken out of a freezer, it must be left at room temperature and allowed to partially melt. "Dry" each ice chunk with a towel before adding to the calorimeter so that you are adding only solid, no liquid. Your collected data should look similar to the time-temperature curve above. CALCULATIONS Specific Heat of water at room temperature is... 4.180 j/g °C The heat gained or lost by water is es (4.180 j/g °C) (temperature change) (mass) Experimental Data 7.87 g Mass of empty calorimeter and stirrer. 67.56 g Mass of calorimeter, stirrer and hot water................. Mass of calorimeter, stirrer and water at the end 94.30 g

Explanation / Answer

(a)

Mass of hot water used = 67.56-7.87 = 59.69 g

(b)

Mass of ice added = 94.3-67.56 = 26.74 g

(c)

Temperature decrease of hot water = 73.21-26.62 = 46.59 0C

(d)

Temperature increase of melted ice = 26.62-0 = 26.62 0C

(e)

Heat lost by hot water = 59.69*4.18*46.59 = 11624.4 J

(f)

Heat lost by calorimeter = 1.317*46.59 = 61.36 J

(g)

Heat gained by melted ice water = 11624.4 + 61.36 = 11685.76 J

(h)

Heat of fusion per gram of ice = 11685.76/(26.74*26.62) = 16.416 J/(g. 0C)

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