I\'m really struggling with these two questions. l. A piece of specific heat \'c

Question

I'm really struggling with these two questions.

l. A piece of specific heat 'cly with a mass of su g is heated waler bath until it reaches a temperature of9R2-c The is transferred into a meter (nested styro dal eups) 68.5314 water initially at 22 am The final temperature of the entire system is 27.5 speci heat capacity of water fie 4,184 J g TC i. Based on this information, determine the value of the calorimeter constant, 2. When 60.0 mL of a 0.400 M ution of HNOs is combined with 60.0 mL of a 0.400 M solution of NaoH in the calorimeter described in question ni. the final temperature of the solution is measured to be 26.6 C. The initial temperature of the solutions is 24.0 c. Assuming the specific heat capacity of the final solution is 3.90 J g i C and the density of the solution is 1.04 g/mL calculate Hint: start with qoolre t goal -0. HNOs (aq) t NaOH

Explanation / Answer

mass of iron = 54.3218 gm

specific heat of iron = 0.452 J/g/oC

Initial temp =98.2oc and final temperature =27.5 oC

heat released by the iron block = 54.3218 gm * 0.452 J/g/oC * (98.2-27.5)oC = 1735.93 J

This amount og heat will be absorbed by the calorimeter + the water.

heat absorbed by the water = 68.5314 gm * 4.184 J/gm/oC * (27.5-22.1)oC =1548.37 J

Heat absorbed by the calorimeter = heat released by the iron blok - heat absorbed by water

= 1735.93 J-1548.37 J = 187.56 J

Change in temperature = 27.5-22.1 = 5.4 oC

Calorimeter constant = 187.56J/5.4oC = 34.73 J/oC

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