Question

Calculate the specific heat (J/g∘C) for a 18.5-g sample of tin that absorbs 183 J when temperature increases from 35.0 ∘C to 78.6 ∘C.

Answer 1

m=18,5g
Q=183J
ΔT=78,6°C-35°C=43,6°C

Q=cmΔT

$c=\frac{Q}{m\Delta T}=\frac{183J}{18,5g*43,6^{o}C}\approx0,227\frac{J}{g^{o}C}}$