How much heat is needed to convert 200 grams of ice at 0 degrees celsius to steam at 100 degrees celsius?
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Answer: 144 kcal
Explanation: For first process (Conversion of ice into water)
Q = mL
= 200 × 80
= 16000 cal
For second process (Rising temperature of water to 100°C)
Q = mcT
= 200 ×1 ×100
= 20000 cal
For thied process (Conversion to steam)
Q = mL
= 200 × 540
= 108000
Q = 16000 + 20000 + 108000
= 144 kcal
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