Gravity pulling the mass of the atmosphere toward the center of the
earth. Atmosphere includes air, water vapor, etc.
<span>2.28 kg x 213 kJ/kg = 486 kJ = 4.86E+05 J</span>
Answer:
Q= mc∆T(ice) + mLF(ice) + mc∆T(water) + mLV(water) + mc∆T(steam)
m=250 g = 0.25 kg = ¼ kg c(ice)= 2100 J/kg.K c(water)= 4200 J/kg.K LF(ice)= 333.7 kJ/kg LV(water)= 2256 kJ/kg c(steam)= 2080 J/kg.K
Explanation:
Q= ¼ × 2100 × (0°-(-30°)) + ¼ × 333700 + ¼ × 4200 × (100°-0°) + ¼ × 2256000 + ¼ × 2080 × (145°-100°)
Q= 15750 + 83425 + 105000 + 564000 + 23400
Q= 791575 J
Answer:
4046atm
Explanation:
For this question you can use the ideal gas law,
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Where P is pressure, V is volume, n is moles of substance, R is the constant, and T is the temperature.
Because of the units given, R will equal .08026
<h3>Rearrange the equation to solve for pressure:</h3>

Then, plug in the values (I'll be excluding units for simplicity, but they all cancel out for pressure in atm):

This will give you:
