Help With Basic Chem

Question

A 59.90g sample of a substance is initially at 27.0 C.  After absorbing 1931 J of heat, the temperature of the substance is 157.9 C. What is the specific heat (c) of the substance?

C=

Answer 1

q = mC∆T

q = heat = 1931 J

m = mass = 59.90 g

C = specific heat = ?

∆T = change in temperature = 157.9º - 27.0º = 130.9ºC

Solving for C...

C = q/m∆T = 1931 J/(59.90 g)(130.9º)

C = 0.24627 J/gºC = 0.2463 J/gºC (to 4 sig. figs.)