1. Put about 200 70°C. Thi mI. of distilled water into a 250-mL beaker, set it o

Question

1. Put about 200 70°C. Thi mI. of distilled water into a 250-mL beaker, set it on the hot plate and heat to about s warm water will be needed for both Part A and Part B. Procedure-Part A 2. Weigh and 3. Add record the mass of two, dry, small, nested foam cups. (Do not tare the cups.) about 60 g of warm water to the nested cups. Weigh and record. Place the nested cups with water in a 250-mL beaker for added stability 4. Work quickly so that the water does not cool too much, record the temperature of the water in the cups. At the same time dry two or three ice cubes (or an amount recommended by the instructor) with paper towels, then add them to the water in the cups. Gently stir with the thermometer until the ice melts. Record the temperature. 6. Remove the cups and contents from the beaker. Weigh and record. 7. Empty and dry the cups. Do not throw them away unless instructed to do so. Procedure-Part B 8. Nest two dry, large foam cups. Label the cups "w" for water. Weigh and record the mass of the empty nested cups. (Do not tare.) 9. Add about 70 g of warm water to the cups. Record the mass of the cups with the warm water. Place the cups and contents in a 400-mL beaker for added stability 10. Nest two more large foam cups. Label the cups "LN" for liquid nitrogen. CAUTION! Do NOT allow liquid nitrogen to come in contact with your skin! Tare these nested cups and add about 40 g of liquid nitrogen. The nitrogen vaporizes rapidly so you will see the mass dropping before your very eyes. Therefore, don't record the mass until the moment you are ready to transfer the liquid nitrogen into the warm water in cups w Lab 13 144

Explanation / Answer

Ans. #1. The temperature of melting ice (or, melting point of ice) = 0.00C

#2. We use nested foam cup to achieve better insulation or to minimize heat change between the system (content of cup) and the surroundings.

#3. Please specify part “B”.

#4. Part A: Given, dHfus of water = 80.4 cal/ g                                        ; [1 cal = 4.184 J]

                                                = (80.4 x 4.184 J) / g

                                                = 336.3936 J/ g

# Molar mass of water = 18.0 g/ mol

So, 1.0 g = (1 / 18.0) mol

Now,

            Molar heat of fusion = dHfus of water / Amount of water in moles

                                                = 80.4 cal/ g              - given

                                                = 80.4 cal / (1/18 mol)

                                                = 1447.2 cal/ mol

                                                = 6055.08 J/ mol

# Part B: Given, dHvap of N2= 47.6 cal/ g                                     ; [1 cal = 4.184 J]

                                                = (47.6 x 4.184 J) / g

                                                = 199.1584 J/ g

# Molar mass of N2 = 28.0 g/ mol

So, 1.0 g = (1 / 28.0) mol

Now,

            Molar heat of vap = dHvap of N2 / Amount of N2 in moles

                                                = 47.6 cal/ g              - given

                                                = 47.6 cal / (1/28 mol)

                                                = 1332.8 cal/ mol

                                                = 5576.44 J/ mol

#5. A, B. Required amount of heat = dHfus of water x Mass of ice

                                                = (80.4 cal/ g) x 56.77 g

                                                = 4564.308 cal

                                                = 19097.064672 J

#6. A, B. Required amount of heat = Molar dHvap of N2 x moles of N2

                                                = (1332.8 cal/ mol) x 2.72 moles

                                                = 3625.216 cal

                                                = 15167.903744 J

#7. Boiling point of N2 = -195.80C

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