://cvg.cengagenow.com/ilrvtakeAssignment/ta mL Use the References to access impo

Question

://cvg.cengagenow.com/ilrvtakeAssignment/ta mL Use the References to access important valees if needed for this question. In the laboratory a "coffes 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 silver weighing 18.05 grams and originally at 97.24 °C is dropped into an insulated eup eontaining 81.58 grams of water at 21.85 °c. Te The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.54 JC Using the accepted value for the specific heat of silver (See the References tool), calculate the final temperature of the water. Assume that no heat is lost to the surroundings Fierious Save and Ex

Explanation / Answer

Let the final temperature be T degree celcius.

heat absorbed =mS(Tf-Ti),m=mass;S=specific heat capacity; Tf=final temperature;Ti=initial temperature

{specific heat of silver=0.240 J/g-C;Water=4.179J/g-C}

heat absorbed by the chunk of silver=18.05*0.240*(T-97.24)

mass of calorimeter=mass of water=21.85 degree celcius

heat absorbed by the calorimeter=heat capacity*change in temperature=1.54 * (T-21.85)

heat absorbed by water=81.58 * 4.179 * (T - 21.85)

according to the law of calorimetry:-

total heat absorbed =total heat released

since all the components is assumed to have absorbed energy,heat released is zero.

18.05*0.240*(T-97.24) + 1.54 * (T-21.85) + 81.58 * 4.179 * (T - 21.85) = 0

4.332T - 421.24368 + 1.54 T - 33.649 + 340.92282T -7449.163617=0

346.79482 T =7904.056297

T(final)=22.7917369 degree celcius.

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