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=(Q)/(m\Delta T)=(183J)/(18,5g*43,6^(o)C)\approx0,227(J)/(g^(o)C)}`