Question

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 63.04 grams of silver to 98.96 °C and then drops it into a cup containing 83.17 grams of water at 22.31 °C. She measures the final temperature to be 25.59 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.61 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of silver Specific Heat (Ag) = J/gºC. In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. Since the cup itself can absorb energy, a separate experiment is needed to determine the heat capacity of the calorimeter. This is known as calibrating the calorimeter and the value determined is called the calorimeter constant. One way to do this is to use a common metal of known heat capacity. In the laboratory a student heats 98.72 grams of nickel to 98.18 °C and then drops it into a cup containing 77.16 grams of water at 23.52 °C. She measures the final temperature to be 32.41 °C. Using the accepted value for the specific heat of nickel (See the References tool), calculate the calorimeter constant. Calorimeter Constant = J°C.

Answer 1

Solution :-

Part 1) mass of water =83.17 g

Mass of silver = 63.04 g

Temperature change of water = 25.59C – 22.31 C = 3.28 C

Temperature change of silver = 98.96 C – 25.59 C = 73.37 C

Specific heat capacity of calorimeter = 1.61 J/C

Step 1) Calculating the heat absorbed by the calorimeter and water

q (water + cal) = (m*c*delta T) + (c cal*Delta T)

                           =(83.17 g * 4.184 J/gC * 3.28 C)+(1.61 J/C * 3.28 C)

                           =1147 J

Heat given off by silver is same as heat absorbed by calorimeter and water

q cal = 1147 J

calculating the specific heat of silver

q= m*c*delta T

c= q/ (m*delta T)

   = 1147 J / (63.04 g * 73.37 C)

   =0.248 J/gC

Therefore the specific heat of silver is 0.248 J/g^oC

Part 2) mass of water =77.16 g

Temperature change of water = 32.41 C- 23.52 C =8.89 C

Mass of nickel =98.72 g

Temperature change of nickel = 98.18 C – 32.41 C = 65.77 C

Calculating the heat absorbed by water

q= m*c*delta T

= 77.16 g * 4.184 J/gC * 8.89 C

   = 2870 J

Calculating heat lost by nickel

q= m*c*delta T

=98.72 g * 0.445 J/gC * 65.77 C

=2890 J

Calculating heat absorbed by calorimeter

q cal = q nickel – q water

          =2890 J – 2870 J

          =20.0 J

Calculating specific heat of calorimeter

c cal = q cal / Delta T

          = 20.0 J / (8.89 C)

          =2.25 J/C

Therefore the calorimeter constant is 2.25 J/^oC