A 10.0 g sample of granite initially at 77.0 ∘C is immersed into 26.0 g of water initially at 22.0 ∘C. What is the final temperature of both substances when they reach thermal equilibrium? (For water, Cs=4.18J/g∘C and for granite, Cs=0.790J/g∘C.)

The final temperature at the equilibrium will be 25.73 °C

Explanation:

Step 1: Data given

Mass of granite = 10.0 grams

Initial temperature of granite = 77.0 °C

Mass of water = 26.0 grams

Initial temperature of water = 22.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qgranite= -Qwater

Q =m* c * ΔT

m(granite) * c(granite) * ΔT(granite) = -m(water) * c(water) * ΔT(water)

⇒with m(granite) = the mass of granite = 10.0 grams

⇒with c(granite) = the specific heat of granite = 0.790 J/g°C

⇒with ΔT(granite) = the change of temperature of granite = T2 - T1 = T2 - 77.0 °C

⇒with m(water) = the mass of water = 26.0 grams

⇒with c(water) = the specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 22.0 °C

10.0 * 0.790 * (T2 - 77.0) = -26.0 * 4.18 * (T2 - 22.0 )

7.9 * (T2 - 77.0) = -108.68 (T2 - 22.0 )

7.9 T2 - 608.3 = -108.68T2 + 2390.96

116.58T2 = 2999.26

T2 = 25.73 °C

The final temperature at the equilibrium will be 25.73 °C

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