8. a) convert heat of fusion of ice to its equivalent inBTU/lb. b) How many BTU

Question

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! 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! 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)]                = 8562BTU

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