Heat of Fusion of Ice A. Determine the number of calories and joules lost by the

Question

Heat of Fusion of Ice

A. Determine the number of calories and joules lost by the warm water.

B. Determine the number of calories and joules needed to warm the melted ice.

C. Determine the number of calories and joules required to melt the ice.

D. Calculate the heat of fusion of ice in cal/g and J/g.

Heat of Fusion of Ice

Initial volume of water (Trial 1 &2) = 25 ml, 25 ml

Initial water temperature (Trial 1 & 2) = 57°C, 55°C

Final volume of water (Trial 1 & 2) = 42 ml, 41 ml

Final temperature of water (Trial 1 & 2) = 2°C, 0°C

Mass of ice melted (Trial 1 & 2) = 17g, 16g

Temperature change (Trial 1 & 2)= 55°C

Instruction (Lab):

Determining the Heat of Fusion of Ice

            Half fill a 250 mL beaker with distilled water. Heat the water in the beaker to about 60EC. Pour about 25 mL of the warmed water into your 25 mL graduated cylinder to warm it. After about a minute or two discard the water and refill the graduated cylinder with exactly 25 mL of the warm water and measure its temperature.

            Obtain an empty styrofoam cup and then add about 50 g of crushed ice to the cup. The exact weight of the ice is not important. Pour off any liquid in the cup. Measure the temperature of the warm water in your 25 mL graduated cylinder, record it on your report sheet and then quickly pour the warm water into the styrofoam cup. Stir the ice-water mixture continuously until the temperature reaches 2EC or less. Record this temperature on your report sheet and immediately pour off the liquid in the styrofoam cup into your 100 mL graduated cylinder. Record this volume on your report sheet. Repeat this process and have your instructor approve your data. If need be, do a third determination.

Explanation / Answer

Trial 1

A) initial temperature of water = 57°C

Final temperature = 2 °C

change in temperture = 55°C

Heat lost by water = Mass of water X change in temperature X specific heat of water

Heat lost = 25 X 55 X 4.18 = 5747.5 Joules

1 joule = 0.239 Cal

5747.5 joule = 0.239 X 5747.5 calories = 1373.65 cal

B. Determine the number of calories and joules needed to warm the melted ice.

The mass of ice melted = 17 grams

Heat required to raise the temperature of ice from 0 to 2 C = Mass X change in temperature X specific heat of water

Heat = 17 X 2 X 4.18 = 142.12 Joules = 142.12 X 0.239 cal = 33.97 Cal

C) The heat lost by warm water = Heat of fusion of ice + heat required to raise the temperature of ice

So Heat required to melt the ice = 1373.65 cal - 33.97 = 1339.68 Cal

Heat required to melt the ice = 5747.5 Joules - 142.12 = 5605.38 Joules

d) Heat of fusion of ice

Heat required to melt the ice / Mass of ice = 1339.68 cal / 17 = 78.80 cal / g

Heat required to melt the ice / Mass of ice = 5605.38 J / 17 g = 329.72 J / g

Trial 2

initial temperature of water = 55°C

Final temperature = 0 °C

change in temperture = 55°C

Heat lost by water = Mass of water X change in temperature X specific heat of water

Heat lost = 25 X 55 X 4.18 = 5747.5 Joules

1 joule = 0.239 Cal

5747.5 joule = 0.239 X 5747.5 calories = 1373.65 cal

B. Determine the number of calories and joules needed to warm the melted ice.

The mass of ice melted = 16 grams

Heat required to raise the temperature of ice from 0 to 0 C = cal

C) The heat lost by warm water = Heat of fusion of ice + heat required to raise the temperature of ice

So Heat required to melt the ice = 1373.65 cal

Heat required to melt the ice = 5747.5 Joules

Heat required to melt the ice / Mass of ice = 1373.65 cal / 16 = 85.85 cal / g

Heat required to melt the ice / Mass of ice = 5747.5 J / 16 g = 359.21 J / g

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