Question

Determine the specific heat of a material if a 35g sample absorbed 48 K as it was heated from 293 k to 313 k.

Answer 1

You have a typo in your assignment "48 K". If there should be 48 J, so here is the solution. If not, then write to me

$Cp = Q/(m\Delta T)$

$\Delta T=313K-293K =20K$

m = 35g = 0,035kg

Q = 48J

$Cp= 48J/(0,035kg*20K)=$ 68,6 J/(kg*K)