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2 July, 14:44

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.)

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  1. 2 July, 17:31
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    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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