How many calories are required to raise the temperature of a 11.6 g sample of iron from 22.2°C to 41.6°C?

1 Answer
Dec 20, 2017

24.2 calories

Explanation:

Using the equation:

E = m xx C xx DeltaT

Where E = energy (J); C = specific heat capacity (Jg^(-1@)C^(-1)); DeltaT = change in temperature (.^@C)

For iron:
C = 0.45 Jg^(-1@)C^(-1)

DeltaT = 41.6 - 22.2
DeltaT= 19.4 ^@C

E = 11.6g xx 0.45Jg^(-1@)C^(-1) xx 19.4^@C
E = 101.3 J

We have our answer in joules, but we need to convert this to calories.

1 cal = 4.184 J

101.3 / 4.184 = 24.2 cal