Question

6.21) Is the specific heat capacity of a substance an intensive or extensive property? Explain.

6.23) Calculate q when 0.10 g of ice is cooled from 10. °C to -75. °C. (S ice = 2.087 J/g•°C).

6.25) A 27.7 g sample of ethylene glycol, a car radiator coolant, loses 688 J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5 °C? (s of ethylene glycol = 2.42 J/g•°C)

6.27) One piece of copper jewelry at 105 °C has exactly twice the mass of another piece, which is at 45 °C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter? (s of copper = 0.387 J/g•°C)

6.29) An unknown volume of water at 18.2 °C is added to 24.4 mL of water at 35.0 °C. If the final temperature is 23.5 °C, what was the unknown volume? (Assume that no heat is lost to the surroundings; d

Pre-lab: Answer the following questions on a separate sheet of paper. To be turned in at the beginning of next week's lab. Calorimetry: Laboratory Measurement of Heats of Reaction (Sample Problems 6.3 -6.5 from the textbook) 6.21 Is the specific heat capacity of a substance an intensive or extensive property? Explain. 6.23 Calculate q when 0.10 g of ice is cooled from 10. °C to -75. °C. (Sice = 2.087 J/g•°C). 5.25 A 27.7 g sample of ethylene glycol, a car radiator coolant, loses 688 J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5 °C? (s of ethylene glycol = 2.42 J/g•°C) 27 One piece of copper jewelry at 105 °C has exactly twice the mass of another piece, which is at 45 °C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter? (s of copper = 0.387 J/g•°C) -9 An unknown volume of water at 18.2 °C is added to 24.4 mL of water at 35.0 °C. If the final temperature is 23.5 °C, what was the unknown volume? (Assume that no heat is lost to the surroundings; d of water = 1.00 g/mL)

of water = 1.00 g/mL)

Answer 1

1.Specific heat capacity of a substance is an intensive property i.e doesn't depend upon amount of substance.Specific heat capacity of a substance is defined as  the amount of heat required to raise the temperature of one gram of that substance by one degree celsius.From this definition it is clear that specific heat capacity doen't depend upon the amount of substance.

2.q=mC$\Delta T$

m=mass of the substance C=Specific heat of the substance $\Delta T$=Tf-Ti=final temperature-initial temperature

here mass of ice=0.1g C=2.087 J/g•°C  $\Delta T$=-75-10=-85°C

q=0.1*2.087*-85=-17.73J

3.HERE q=688J mass=27.7 g Final temperature=32.5°C Initial temperature=Ti sp.heat capacity =2.42J/g•°C  

q=mC$\Delta T$=mC(Tf-Ti)

688=27.7*2.42*(32.5-Ti)

by solving the above we get=Ti=22.236°C

For 4 and 5 please find the attached image

4. Heat loss & Heat gain a= mcoT Gleat loss : 2m x cx (105-T) dit I final tembereature. (= specific heat of copper m = Mass of Anothere piece of coppere oleat gain = mx cx (T-45) Then and xf x (105 - 1) = x 8 x (T-45) => 9. T-45 105-T => 210-27 = T- us - 210tus : 37 - 255 = 31 = 7 T = 85°c S. Let the unknow Mass of water = ng known Mass of watere & Density & Volume = 19/mex 24.4 DAL F 24.49 Heat loss - Heat gain 9 24.4 X XC35-23.5 °C = **&* ( 23.5-18-2°c c. specific heat of water To 23.5 CS Scanned with ux ihned with 4 x 11.5 = xx 5.3 = x = 52.94g a Volume = Mass 52.948 = 52.94m Density Tolme CamScanner-