How many grams of gasoline would you need to burn if you started with a 275 gram piece of ice that started at -25 OC and by the end of the experiment had 100 grams of steam and the rest liquid water?
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This is a many step problem. How much have you done and exactly what do you not understand.
CpxmassxTf-Ti
(2.06)(275)(0-(-25))=14162.5
333J/100x14162.5 = 47161.125
47161.125+14162.5= 61323.625
61323.625(1gas/48,000)= answer: 1.277g of steam
Thank you to the way for helping chemistry isn't my best subject so I apreciate the time y'all put in to help tutor
CpxmassxTf-Ti
(2.06)(275)(0-(-25))=14162.5 This step is perfect. Good work!
333J/100x14162.5 = 47161.125 I don't understand this step. You should be melting all 275 g of the ice so it should be 333 J/g x 275 g = ? J for the melting step.
Now you want to raise the temperature of the 275 g water from zero C to 100 C which is 275g x 4.184 J/g x (100-0) = ?
Then you want to vaporize 100 g of the 275 g H2O you have to steam which will leave you with 175 g of liquid water.
That is 100 g H2O x heat vaporization which I think is about 2260 J/g (but check me out on that for sometimes I don't remember all of these constants) = ? J.
All all of those values together to find total q needed.
Then look up the heat of combustion of gasoline and we can talk about that step when you get there.
47161.125+14162.5= 61323.625
61323.625(1gas/48,000)= answer: 1.277g of steam
To determine the grams of gasoline needed to burn in order to achieve the described experiment, we need to calculate the heat required to convert the ice to steam and the liquid water to steam. This calculation involves two steps:
Step 1: Calculate the heat required to convert the ice to liquid water.
The heat required is given by the equation:
Q1 = m1 × c1 × deltaT1,
where:
- Q1 represents the heat (in joules) required to convert the ice to liquid water,
- m1 is the mass of the ice (275 grams),
- c1 is the specific heat capacity of ice (2.09 J/g°C), and
- deltaT1 is the change in temperature from -25°C to 0°C (which is 25°C).
Using the given values:
Q1 = 275 g × 2.09 J/g°C × 25°C.
Step 2: Calculate the heat required to convert the liquid water to steam.
The heat required is given by the equation:
Q2 = m2 × c2 × deltaT2,
where:
- Q2 represents the heat (in joules) required to convert the liquid water to steam,
- m2 is the mass of the water (275 grams - 100 grams = 175 grams),
- c2 is the specific heat capacity of liquid water (4.18 J/g°C), and
- deltaT2 is the change in temperature from 0°C to 100°C (which is 100°C).
Using the given values:
Q2 = 175 g × 4.18 J/g°C × 100°C.
Finally, to determine the total heat required for the experiment, add Q1 and Q2:
Q_total = Q1 + Q2.
Please note that the specific heat capacity values provided are approximate and may vary slightly depending on the source.