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2 years ago

PLEASE SHOW YOUR WORKCalculate how much energy would be necessary to convert 250.0 grams of ice at -30°C to steam at 145°C.

Chemistry

1 answer:

garri49 [273]2 years ago

7 0

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

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**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.

PLEASE SHOW YOUR WORKCalculate how much energy would be necessary to convert 250.0 grams of ice at -30°C to steam at 145°C.