A 22.36 g sample of a substance is initially at 27.7∘C. After absorbing 1319 J of heat, the temperature of the substance is 148.0 ∘C. What is the specific heat capacity of the substance?

Question

0.490 is incorrect,0.49 incorrect, 0.5 incorrect.

Answer 1

q = mC∆T

q = heat = 1319 J

m = mass = 22.36 g

C = specific heat = ?

∆T = change in temp = 120.3º

C = q / m∆T = 1319 J / (22.36 g)(120.3º)

C = 0.4904 J/gº

If they are picky about sig. figs. they may want the answer to 4 sig. figs. even though the temperature has only 3 sig.figs. Otherwise, the answer key is wrong, because this is the correct answer.