A12.5 gram sample of granite initially at 82.0 oc is immersed into 25.0 grams of water initially at 22.0 oc. what is the final temperature of both substances when they reach thermal equilibrium? (for water, cs = 4.18 j/g oc and for granite, cs = 0.790 j/g oc

Let's use the concept of conservation of energy.

Q for granite = Q for water
(mass of granite)(Cs of granite)(ΔT) = (mass of water)(Cs of water)(ΔT)
where ΔT is the difference of the initial temperature, T₀, and the equilibrium temperature T. T would be between 82°C and 22°C. Make sure that ΔT is positive.

T₀ for granite = 82°C
T₀ for water = 22°C

Substituting the values and solving for T,
(12.5)(0.79)(82 - T) = (25)(4.18)(T - 22) 
T = 27.18°C