Heatavail= warmingice + melting ice + heating water
441.8Kj=m*cice*(0-(5.2C))+Hm*m+m*cwater*(6.4-0)
solve for m.
HOW MANY GRAMS OF ICE AT -5.2 DEGREES CELSIUS CAN BE COMPLETELY CONVERTED TO LIQUID AT 6.4 CELSIUS DEGREES IF THE AVAILABLE HEAT FOR THE PROCESS IS 441800kj,FOR ICE SPECIFIC HEAT IS 2.01
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