Adding 1.53x10^3J of electrical energy to a constant-pressure calorimeter change

Question

Adding 1.53x10^3J of electrical energy to a constant-pressure calorimeter changes the water temperature from 30.50 degrees Celsius to 31.85 degrees Celcius. When 1.75g of a solid salt is dissolved in the water, the temperature falls from 21.85 to 21.44 degrees Celsius. Find the value of q for the solution process. Adding 1.53x10^3J of electrical energy to a constant-pressure calorimeter changes the water temperature from 30.50 degrees Celsius to 31.85 degrees Celcius. When 1.75g of a solid salt is dissolved in the water, the temperature falls from 21.85 to 21.44 degrees Celsius. Find the value of q for the solution process.

Explanation / Answer

When 1.53*103 joules of energy is added, this results in rise of mass of water temperature from 30.50 to 31.85 deg.c. Let m= mass of water, specific heat of water= 4.184 J/gm.deg.c

heat supplied = heat supplied to water= mass of water* specific heat of water* change in temperature

1.53*1000 = m* 4.184* ( 31.85-30.50)

m= 271 gm

when 1.5 gm of salit is added, total mass of solution = mass of water+ mass of salt= 271+1.75= 272.75 gm. The addition of mass of salt resulted in drop in solution temperature from 21.85 to 21.44 deg.c suggesting the process to be endothermic where heat is absorbed. Assuming the solution to be dilute which enables the solution specific heat to be same as that of water,

heat removed, q= mass of solution* specific heat of water* change in temperature

=272.75*4.184*(21.85-21.44) joules=513.5 joules

this much heat is absorbed.

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