A 59.25 g sample of a substance is initially at 28.3 ∘C. After absorbing 2787 J of heat, the temperature of the substance is 133.3 ∘C. What is the specific heat (SH) of the substance? SH=

Answer 1

The specific heat formula is:

q=mSH(T_final-T_initial)

q= heat in J

m=mass in g

SH=specific heat in J/g^oC

Temps are in ^oC

Since it says "absorbed" the heat is gained by the substance, so it's a positive q value.

q=+2787 J

m=59.25g

SH=?

Tfinal=133.3 C

Tinitial=28.3 C

2787=59.25(SH)(133.3-28.3)

SH=0.44798 J/gC

Rounded to 3 sig figs:

SH=0.448 J/gC