How many grams of liquid water can be solidified by the release of 0.75kJ of heat?
Would the first step be (0.75 kJ)(1 mol/6.01 kJ) ??
Thanks
Heat=Heatfusion*mass
mass= Heatavailable/heatfusion
Well, well, well, trying to solidify some water, are we? Alright, let's see.
To figure out how much water can be solidified by the release of 0.75 kJ of heat, you'll need to know the enthalpy of fusion of water.
The enthalpy of fusion of water is about 334 J/g (or 334 kJ/kg, if you prefer). So, you can use this magical conversion factor to help you out.
If you divide 0.75 kJ by 334 kJ/kg, you'll be left with the number of kilograms that can be solidified. Then, you can use the fact that 1 kilogram of water is equal to 1000 grams to find the number of grams.
So, you're on the right track with the setup. Just remember to put the conversion factor in the right place, and the answer shall be revealed!
Yes, the first step would be to convert the given amount of heat (0.75 kJ) into moles of water. To do this, you can use the molar heat capacity of water.
The molar heat capacity of water is approximately 4.18 J/(g·°C), which means it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Since the heat released (0.75 kJ) is given in kilojoules, you need to convert it to joules by multiplying by 1000. So, 0.75 kJ = 0.75 × 1000 J = 750 J.
Now, divide the heat energy in joules (750 J) by the molar heat capacity (4.18 J/(g·°C)) to get the number of grams of water that can be solidified:
(750 J) / (4.18 J/(g·°C)) = 179.43 grams
Therefore, approximately 179.43 grams of liquid water can be solidified by the release of 0.75 kJ of heat.