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

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 chunk of gold weighing 19.09 grams and originally at 97.99 °C is dropped into an insulated cup containing 81.00 grams of water at 22.78 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.90 J/°C.

Using the accepted value for the specific heat of gold, .129 J/goC, calculate the final temperature of the water. Assume that no heat is lost to the surroundings.

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 19.09 grams and originally at 97.99 °C is dropped into an insulated cup containing 81.00 grams of water at 22.78 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.90 J/°C.

Using the accepted value for the specific heat of gold, .129 J/goC, calculate the final temperature of the water. Assume that no heat is lost to the surroundings.

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. Thermometer A chunk of gold weighing 19.09 grams and originally at 97.99 °C is dropped into an insulated cup containing 81.00 grams of water at 22.78 °C. Stirring rod The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.90J/C Water Metal sample 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 final

Explanation / Answer

mass of gold =mgold 19.09 g

Initial temperature =T1 = 97.99 °C

mass of water = 81 g

Water Initial temperature = T1 = 22.78 °C

Specific heat capacity of glod = Cgold = 0.129 J/g.°C

Specific heat capacity of water= CH2O = 4.18 J/g.°C

specific heat capacity of cup = C cup = 1.90 J/°C

Final temperature of gold = water = cup = T2

Heat lost by the gold = heat gained by the water + heat gained by the calorimeter

Qgold = QH2O + Qcalorimeter

mC(T1-T2) = mC(T2-T1) + C(T2-T1)

19.09(0.129)(97.99-T2) = 81(4.18)(T2-22.78)* (1.90)(T2-22.78)

2.463(97.99-T2) = 338.58(T2-22.78)2 * (1.90)

2.463(97.99-T2) = 643.3(T2-22.78)2

97.99-T2 = 261.2(T2-22.78)2

By solving this T2 = 23.3°C

So final temperature of water = 23.3 °C

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