7500. mg of a metal absorbs 18.7cal as it is heated from 33 to 99C. What is the specific heat in cal/goC Answer in appropriate amount of significant figures.

Question

1. 7500. mg of a metal absorbs 18.7cal as it is heated from 33 to 99C.  What is the specific heat in cal/g^o_C  Answer in appropriate amount of significant figures.

Answer 1

q = mC∆T

q = heat = 18.7 cal

m = mass = 7500. mg x 1 g / 1000 mg = 7.500 g

C = specific heat = ?

∆T = change in temperature = 99º - 33º = 66º

Solving for C ..

C = q / (m)(∆T) = 18.7 cal / (7.5000 g)(66º)

C = 0.03778 cal / gº = 0.038 cal / gº (2 sig.figs.)