Question

Question #42d33

Answer 1

The food caloric content of peanut butter is `"5.3 Cal/g"`.

So, you know the heat capacity of 16 g of peanut butter; in this case, peanut butter's heat capacity expresses the heat given off per degree Celsius by your sample of peanut butter.

What you need to do is determine the total amount of heat given off that corresponds to your respective change in temperature, `DeltaT`

`q = C * DeltaT = 110 "kJ"/(cancel(@"C")) * (25.3 - 22.1)^@cancel("C")`

`q = "352 kJ"`

The next thing you need to determine is how much heat is given off per gram, since that much heat was given off by 16 g

`q = "352 kJ"/"16 g" = "22 kJ/g"`

From this point on, all you have to do is perform a series of unit conversions to go from kJ per gram to J per gram, then from J per gram to calories per gram, then from calories per gram to kilocalories, or "large calories", per gram

`22cancel("kJ") * (1000cancel("J"))/(1cancel("kJ")) * (1 cancel("calorie"))/(4.184cancel("J")) * "1 Cal"/(1000cancel("calories")) = "5.258 Cal/g"`

Rounded to two sig figs, the answer will be

`q = color(green)("5.3 Cal/g")`