A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final temperature of both substances when they reach thermal equilibrium? (For water, Cs=4.18J/g⋅∘CCs=4.18J/g⋅∘C and for granite, Cs=0.790J/g⋅∘CCs=0.790J/g⋅∘C.)

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °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

Heat lost = heat gained

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 = 14.4 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 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

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

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

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