How much heat in joules is required to heat a 43g sample of aluminum from an initial temperature of 72˚f to a final temperature of 145˚f ? special heat capacity for aluminum is 0.903 j/g˚c

The first step to answering this item is to convert the given temperatures in °F to °C through the equation,

   °C = (°F - 32)(5/9)

initial temperature: 72°F

    °C = (72 - 32)(5/9) = 22.22°C

final temperature: 145°F
   
    °C = (145 - 32)(5/9) = 62.78°C

Substituting to the equation,

    H = mcpdT
    
    H = (43 g)(0.903 J/g°C)(62.78 - 22.22)
   
   H = 1574.82 J

1574.82 J

Sorry i think the answer is b