A sample of gold absorbs 34.6 J of heat as its temperature is increased by 62.0 C. What is the mass of the gold sample? The specific heat of gold is 0.129 j/gx C

Answer 1

q = mC∆T

q = heat = 34.6 J

m = mass = ?

C = specific heat = 0.129 J/gºC

∆T = change in temperature = 62.0ºC

Solving for m (mass) we have...

m = q/(C)(∆T) = (34.6)/(0.129)(62)

m = 4.33 g