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 tin weighing 19.32 grams and originally at 97.28 °C is dropped into an insulated cup containing 75.94 grams of water at 23.32 °C.  



The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.52 J/°C.



Using the accepted value for the specific heat of tin (See the References tool), calculate the final temperature of the water. Assume that no heat is lost to the surroundings.



Tfinal =  °C.

Explanation / Answer

Heat lost by metal = heat gained by water + heat gained by cup

Heat lost be metal = Q = mcT

m = mass of tin = 19.32 grams

Initial temperature = T1 = 97.28°C

Let final temperature = T2 = ?

heat capacity of tin = 0.21 J / g °C

So Q = heat loss by tin = 19.32 X 0.21 X (97.28 - T2)

Q1 = Heat gained by water = Mass of water x specific heat of water X change in temperature

Q1 = 75.94 X 4.2J/(gC). X (T2 - 23.32)

Q2 = Heat gained by calorimeter = Heat capacity X change in temperature = 1.52 X (T2- 23.32)

So

Q = Q1+ Q2

19.32 X 0.21 X (97.28 - T2) = 75.94 X 4.2J/(gC). X (T2 - 23.32) + 1.52 X (T2- 23.32)

4.05 (97.28 - T2) = (318.94 + 1.52 ) ( T2-23.32)

393.98 - 4.05T2 = 320.46 T2 - 7473.12

7867.1 = 324.51 T2

T2 = Final temperature = 24.24 oC

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