PLEASE ANSWER G-I Answers given above are incorrect for G-I please answer correc

Question

PLEASE ANSWER G-I
Answers given above are incorrect for G-I please answer correctly BEFORE you can begin the Heat of Fusion of Ice experiment you MUST: 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 Time T Add the required amount of hot (60-70°C) water Curve (Logger-Pro Ice is being added and reweigh. Start stirring. Insert a temperature sensor and "continuously" record the temperature vs time. H of hot water. After about 30 seconds start adding pieces of ice until the temperature drops to between 20 and Final temperature of water 30°C. Record this final temperature for at least 30 seconds. Make sure there is efficient stirring and all the ice has melted. Finally weigh the calorimeter and final contents. NOTE: the ice must be at oec so, if taken out of a freezer, it must be len atroom 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 jg °C. The heat gained or lost by water is joules (4.180 jg °C (temperature change)(mass) Experimental Data Mass of empty calorimeter and stirrer. 787 g Mass of calorimeter, stirrer and hot water 67.56 g Mass of calorimeter, stirrer and water at the end 94.30 g Calorimeter Constant. 317 iMPC

Explanation / Answer

SOLUTION:

(A) MAss of hot water = 67.56g - 7.87g = 59.69g

(B) Mass of ice = 94.30g - 67.56g = 26.74g

(G) Heat gained by the melted ice water = Mice X SH2O X (Tf - Ti)

Mice = mass of ice = 26.74g; SH2O = specific heat of water = 4.180 j / goC ; Tf = 26.62 oC; Ti = 0 oC

Heat gained by the melted ice water = 26.74 X 4.180 j / goC X (26.62 oC - 0 oC)

Heat gained by the melted ice water = 2977. 63 j

(H) Use the equation

Hea lost by water = heat gained by water + heat gained by calorimeter

Mwater X SH2O X (Ti - Tf) = Heat used to melt the ice + heat used to raise the temp of ice + Scal X (Ti - Tf)

Scal = specific heat of calorimeter.

59.69 X 4.180 X (73.21 - 26.62) = (Mice X HL ) + 2977. 63 j + 1.317 j/oC (73.21 - 26.62)

HL = heat of fusion of ice

11624.4j = (26.74 X HL ) + 2977. 63 j + 61.35 j

(26.74 X HL ) = 11624.4j - 2977. 63 j - 61.35 j = 8585.42j

HL =  8585.42j / 26.74g = 321.07 j/g

(I) Entropy change ( S) = HL /T

S = 321.07 j/g / 273K = 1.176 j/g/K

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