    Question

# Calculate the amount of heat required to convert 1.00 kg of ice at −10∘C into steam at 100∘C at normal pressure. (Sice=2100 J kg−1K−1, Swater=4200 J kg−1K−1, Lf,ice=3.36×105 J kg−1, Lv,water=2.25×106 J kg−1)

A

3.03×106 J

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B

4.03×106 J

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C

5.03×106 J

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D

6.03×106 J

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Solution

## The correct option is A 3.03×106 J Heat required to heat ice from −10∘C to ice at 0∘C is: Q1=mSiceΔT1 Q1=1 kg×2100 J kg−1K−1×10 K Q1=21000 J Heat required to covert ice to water is: Q2=mLf,ice Q2=1 kg×3.36×105 J kg−1 Q2=336000 J Heat required to heat water from 0∘C to water at 100∘C is: Q3=mSwaterΔT3 Q3=1 kg×4200 J kg−1k−1×100 Q3=420000 J Heat required to covert water to steam is: Q4=mLv,water Q4=1 kg×2.25×106 J kg−1 Q4=2250000 J Total heat required is: Q=Q1+Q2+Q3+Q4 Q=(21+336+420+2250) kJ Q=3.027×106 J  Suggest Corrections  0      Similar questions
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