500 grams of ice at -10^oC will be melted down to become water at 5^oC. If the specific heat of ice is 0.5 cal/gr^oC, the heat of fusion for ice is 80 cal/gr and the specific heat of water is 1 cal/gr^oC. Determine the amount of heat needed * 20 poin​

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500 grams of ice at -10^oC will be melted down to become water at 5^oC. If the specific heat of ice is 0.5 cal/gr^oC, the heat of fusion for ice is 80 cal/gr and the specific heat of water is 1 cal/gr^oC. Determine the amount of heat needed * 20 poin​

Jawaban:

Penjelasan:

– heat absorbed to reach 0 degree C

Q_1=mcDelta T=500 gramtimes 0.5 cal/gramKtimes (0-(-10))K=2500 cal

– heat absorbed to make it water

Q_2=mL=500gramtimes 80 cal/gram=40000 cal

– heat absorber to reach 5 degree C of water

Q_3=mcDelta T=500 gram times 1 cal/gramKtimes (5-0)K=2500 cal

total heat absorbed

Q=Q_1+Q_2+Q_3=2500cal+40000cal+2500cal=45000cal