Weigh a dry calorimeter containing a magnetic stirring bar. Add the required amo

Question

Weigh a dry calorimeter containing a magnetic stirring bar.

Add the required amount of hot (60-70°C) water and reweigh.

Start stirring. Insert a temperature sensor and "continuously" record the temperature vs time.

After about 30 seconds start adding pieces of ice until the temperature drops to between 20 and 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.

Experimental Data

CALCULATE: (to the correct number of sig figs; use correct units; mass in g (grams), energy in j (joules)).

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.

Add the required amount of hot (60-70°C) water and reweigh.

Start stirring. Insert a temperature sensor and "continuously" record the temperature vs time.

After about 30 seconds start adding pieces of ice until the temperature drops to between 20 and 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.

Heat of Fusion of Ice Time-Temperature Curve (Logger-Pro) — Ice is being added Initial temperature of hot water. Final temperature of water time

Explanation / Answer

(a) Mass of hot water = (82.52-6.25) grams = 76.27 grams

(b) mass of ice added= (98.84-82.52) grams = 16.32 grams

(c) Temperature decrease of hot water = (Final temperature of water in calorimeter - temperature of hot water) = (27.66 0 C - 52.79 0 C) = - 25.13 0 C ( negative sign indicates that temperature is decreasing)

(d) Temperature rise of melted ice = (Final temperature of water in calorimeter - temperature of ice = (27.66 0 C - 00 C) = 27.66 0 C

(e) Heat lost by water: Mass of water * Specific heat of water * Temperature decrease = (76.27 grams)* (4.180 J/grams 0 C)* ( 25.13 0 C) = 8012 Joules

(f) Heat lost by calorimeter = Calorimeter Constant * Temperature decrese = (0.310 J/ 0 C) * (25.13 0 C) = 7.8 Joules

(g) Heat gained by melted ice water = Mass of melted ice water * Specific heat of water * Temperature rise = (16.32 grams)* (4.180 J/ g 0 C) * ( 27.66 0 C) = 1887 J

(h) Heat of fusion per gram of ice = ( 8012-+7.8-1887) J/ 16.32 grams = 376 J

(i) Entropy change of ice @ 273.15 K = Enthalpy of fusion of water per mole/ 273.15 K = (376*18/273.15) J/mole K = 24.78 J/mole K

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