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A 113.25 gram sample of gold is initially at 100.0 °C. It gains 20.00 J of heat from its surroundings. What is its final temperature? (specific heat of gold = 0.129 J g−1 °C−1)

The equation we need that involves heat and change in temperature is

q=mcΔT where q is heat in Joules, m is mass in grams, c is specific heat capacity in J/gC and ΔT is Tfinal -Tinitial.

Let's list everything that is given.

q = 20.00 J

m = 113.25 g

c = 0.129 J g−1 °C−1

Ti = 100.0 °C



20.00 J = 113.25 g*0.129 J g−1 °C−1*ΔT

ΔT = 1.368996 = Tfinal -Tinitial

1.368996 = Tfinal -100.0 °C

Tfinal = 101 °C

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