Answer:
B
Explanation:
Shows actual sacrifice in order to train and be the best.
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK
</span>
Now, energy required for melting of ICE = <span> 334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC = 4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span> 2257X 5.25 = 11849.25 J ......(3)
</span><span>
Thus, total heat energy required for entire process = (1) + (2) + (3)
= 1753.5 + 2195.18 + 11849.25
= </span><span>15797.93 J
</span><span> = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
As water is boiled, kinetic energy causes<span> the </span>hydrogen bonds to break<span> completely and allows water molecules to escape into the air as gas (steam or water vapor). When water freezes, water molecules form a crystalline structure maintained by </span>hydrogen bonding<span>. Solid water, or </span>ice<span>, is less dense than liquid water.</span>
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
