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

Answer: 1574.82 J