I'm trying to figure out the enthalpy of fusion, then compare it to the enthalpy fusion...

I'm trying to figure out the enthalpy of fusion, then compare it to the enthalpy fusion of ice, for an experiment that I did in a class the other day. It's really confusing me! The process involved placing ice into a calorimeter with room temperature water inside of it, then recording the change in temperature over a period of time. Thanks for the help.

Calorimeter mass = 7.96g

Calorimeter sp.heatcapacity = 3.3J/gK

Calorimeter temp change = (21.2C - 13.1C)

Ice mass - 9.58g

Water mass - 97.90g

Water temp change = (21.2C-13.1C)

Solutions

Expert Solution

Heat released by calorimeter : Mass of Calorimeter * Specific heat capacity of Calorimeter * Temperature change = 7.96 g * ( 3.3 J/ g K) * ( 21.2- 13.1) K = 212.8 J

Heat released by room temperature water : Mass of Calorimeter * Specific heat capacity of water * Temperature change = 97.90 g * ( 4.18 J/ g C) * ( 21.2- 13.1) C = 3314.7 J

Total heat released : 212.8 + 3314.7 J = 3527.5 J

Heat gained by ice = heat gained for melting of ice + heat gained by water for the increase in temperature

Heat gained by melting for ice : enthalpy of fusion ( H J/g) * mass = H * 9.58 J

heat gained by water for the increase in temperature = 9.58 g * 4.18 J/ g ⁰C * ( 13.1 - 0 ) ⁰C = 524.6 J

So, total heat gained : 9.58 H + 524.6

At thermal equilibrium, 9.58 H + 524.6 = 3527.5

Or, H = ( 3527.6 -524.5) / 9.58 = 313.5

The enthalpy of fusion of ice is: 313.5 J/ g = 313.5 * 18 J / mole = 5643 J/ mole = 5.643 kJ/ mole

(Molar mass of ice = 18 g/ mole)

The actual value of enthalpy of fusion of ice is 6.01 kJ/ mole

queen_honey_blossom answered 1 year ago

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