A pound I’m unsure but I hope it’s right
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>
i believe that momentum is the answer :)
In order to answer this, we mus know the data for the heat of combustion of propane. This is an empirical data that you can search online. The heat of combustion is -2220 kJ/mol. The molar mass of propane of 44.1 g/mol. The solution is as follows:
ΔH = -2220 kJ/mol (1 mol/44.1 g)(1000g/1kg)(20 kg)
<em>ΔH = -1006802.721 kJ or -1 GJ</em>