In the laboratory a \"coffee cup\" calorimeter, or constant pressure calorimeter

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 63.58 grams of platinum to 98.58 °C and then drops it into a cup containing 80.90 grams of water at 24.09 °C. She measures the final temperature to be 25.98 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.72 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of platinum. Specific Heat (Pt) = J/g°C

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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 93.30 grams of lead to 98.15 °C and then drops it into a cup containing 75.98 grams of water at 22.09 °C. She measures the final temperature to be 25.46 °C. Using the accepted value for the specific heat of lead (See the References tool), calculate the calorimeter constant. Calorimeter Constant = J/°C.

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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 chunk of gold weighing 19.87 grams and originally at 97.12 °C is dropped into an insulated cup containing 84.79 grams of water at 21.39 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.63 J/°C. Using the accepted value for the specific heat of gold (See the References tool), calculate the final temperature of the water. Assume that no heat is lost to the surroundings.

Explanation / Answer

Ans 1

Heat gained by water + heat gained by calorimeter = Heat lost by platinum

Mass of water x specific heat capacity of water x (Tf - Ti) + calorimeter constant x (Tf - Ti) = Mass of platinum x specific heat capacity of platinum x (Tf - T1)

80.90 g x 4.184 J/gC x (25.98 - 24.09)C + 1.72 J/C x (25.98 - 24.09)C = 63.58 g x Cp x ( 98.58 - 25.98)C

642.988584 = 4615.908 x Cp

Specific Heat (Pt), Cp = 0.139 J/gC

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