**PLEASE SHOW YOUR WORK**Calculate how much energy would be necessary to convert 250.0 grams of ice at -30°C to steam at 145°C. Assume that the only time that liquid water changes to gas is when it is boiling. Please help me

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

the answer is d   for sure : )

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation: