1 BTU=252 cal 8. a) convert heat of fusion of ice to its equivalent inBTU/lb. b)
ID: 1740250 • Letter: 1
Question
1 BTU=252 cal 8. a) convert heat of fusion of ice to its equivalent inBTU/lb. b) How many BTU are absorbed by a refrigerator in changing 5.0lbwater at 60degrees F to ice at 32 degrees F? show me all work and Answers so I can check it! 1 BTU=252 cal 8. a) convert heat of fusion of ice to its equivalent inBTU/lb. b) How many BTU are absorbed by a refrigerator in changing 5.0lbwater at 60degrees F to ice at 32 degrees F? show me all work and Answers so I can check it! 1 BTU=252 cal 8. a) convert heat of fusion of ice to its equivalent inBTU/lb. b) How many BTU are absorbed by a refrigerator in changing 5.0lbwater at 60degrees F to ice at 32 degrees F? show me all work and Answers so I can check it! show me all work and Answers so I can check it!Explanation / Answer
(a) We have heat of fusion of ice is L =3.33*105J/kg =(3.33*105J/kg)(1cal/4.186J)(1BTU/252Cal)(1kg/2.205lb) =143.16BTU/lb (b) Now specific heat of water is Cwater =4186J/kg.oC =(4186J/kg.oC)(1cal/4.186J)(1BTU/252Cal)(1kg/2.205lb) =1.8 BTU/lb.oC Here water is initially at 60oF This temperature in celsius scale is equal to (60 -32)(5/9) = 15.6oC Now heat absorbed by the refrigerator in changing 5lb water at60oF to ice at 32oF is Q= mwaterCwaterT + mwaterLice = (50lb)[(1.8BTU/lb.oC)(15.6oC) + (143.16BTU/lb)] = 8562BTU = (50lb)[(1.8BTU/lb.oC)(15.6oC) + (143.16BTU/lb)] = 8562BTURelated Questions
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