Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero degrees Celsius, melts at zero degrees Celsius, and boils at 100 degrees Celsius. Remember that you need to take into account three changes: melting ice, heating water, and vaporizing the water.)

Question

11-05-2018
Chemistry

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Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero degrees Celsius, melts at zero degrees Celsius, and boils at 100 degrees Celsius. Remember that you need to take into account three changes: melting ice, heating water, and vaporizing the water.)

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PBCHEM PBCHEM · Answer 1 · 2018-05-20T06:56:17+02:00

Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g
Heat of vaporization of H2O = 2257 J/g
Heat capacity of H2O = 4.18 J/gK

Now, energy required for melting of ICE =   334 X 5.25 = 1753.5 J .......(1)
Energy required for raising the temperature water from 0 oC to 100 oC =  4.18 X 5.25 X 100 = 2195.18 J .............. (2)
Lastly, energy required for boiling water =   2257X 5.25 = 11849.25 J ......(3)

Thus, total heat energy required for entire process = (1) + (2) + (3)
= 1753.5 + 2195.18 + 11849.25
= 15797.93 J
= 15.8 kJ
Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-04-14T12:16:07+02:00

The amount of heat needed to boil 5.25 grams of ice when we take into account three changes as melting, heating and vaporizing the water is 15797.93 J.

How do we calculate total heat?

Total heat for the given condition will be calculated by the addition of the heat of fusion, heat of vaporization and specific heat of water.

In the question it is given that,

mass of ice = 5.25 grams

Change in temperature = 100 - 0 = 100 degree celsius

For the melting of ice:

We know that heat of fusion of water = 334 J/g

Required heat for the melting of ice = 334 × 5.25 = 1753.5 J

For heating water:

Amount of heat will be used by using the formula as,

Q = mcΔT, where

c = specific heat of water = 4.18 J/gK

Required heat for heating of water = 4.18 × 5.25 × 100 = 2195.18 J

For vaporization of water:

We know that heat of vaporization of water = 2257 J/g

Required heat for vaporization of ice = 2257 × 5.25 = 11849.25 J

Total amount of heat involved = 1753.5 + 2195.18 + 11849.25 = 15797.93 J

Hence total amount of heat is 15797.93 J.

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