The specific heat of mercury is .03cal/g*C, and it's boiling point is 357*C. The specific heat of water is 1 cal/g*C. It takes 65 calories of energy to vaporize one gram of mercury and 540 calories to vaporize one gram of water. If both substances begin at about 22*C does it take more energy to boil a gram of mercury or a gram of water? If someone can give me the formula to use, I can try to figure it out myself! Thanks
heat to boil 1 gram water:
mcdeltaT+ Lv*m
1*1*(100-22)+540*1
heat to boil 1 gram Hg
mcdeltatT+Lv*m
1*.03(357-22)+65*1
do the math.
Energy required per gram =
Q = (heat of vaporization) + (Tboiling - Tinitial)*(specific heat)
Compute and compare that quantity for both liquids.
For mercury,
Q = 65 + 335*0.03
For water
Q = 540 + 78*1.0
Water's heat requirement is much higher
To determine whether it takes more energy to boil a gram of mercury or a gram of water, we can compare the energy required to heat each substance from 22°C to their respective boiling points, and then the energy required to vaporize them.
The formula we can use is:
Energy = Mass * Specific Heat * Change in Temperature
1. First, let's calculate the energy required to heat each substance from 22°C to their boiling points:
For mercury:
Energy = 1g * 0.03 cal/g°C * (357°C - 22°C) = 10.05 cal
For water:
Energy = 1g * 1 cal/g°C * (100°C - 22°C) = 78 cal
2. Next, let's calculate the energy required to vaporize each substance:
For mercury:
Energy = 65 cal
For water:
Energy = 540 cal
3. Finally, let's add the energy required for heating and vaporization for each substance:
For mercury:
Total energy = 10.05 cal + 65 cal = 75.05 cal
For water:
Total energy = 78 cal + 540 cal = 618 cal
Comparing the total energy required, we can see that it takes less energy to boil a gram of mercury (75.05 cal) compared to boiling a gram of water (618 cal).
Therefore, it takes more energy to boil a gram of water than a gram of mercury.
I hope this explanation helps! Let me know if you have any further questions.