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10 October, 14:49

A 1.10 kg piece of granite with a specific heat of 0.803 J g-1 °C-1 and a temperature of 87.3 °C is placed into 2.00 L of water at 20.3 °C. When the granite and water come to the same temperature, what will the temperature be?

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  1. 10 October, 15:07
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    The final temperature will be 26.7 °C

    Explanation:

    Step 1: Data given

    Mass of the piece granite = 1.10 kg

    Specific heat = 0.803 J/g°C

    Initial temperature = 87.3 °C

    Volume of water = 2.00 L

    Initial temperature of water = 20.3°C

    Step 2:Calculate the final temperature

    Heat lost by granite = heat won by water

    Qgranite = - Qwater

    Q = m*c*ΔT

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

    ⇒ mass of granite = 1100 grams

    ⇒ c (granite) = 0.803 J/g°C

    ⇒ ΔT (granite) = T2 - T1 = T2 - 87.3

    ⇒ mass of water = 2000 grams

    ⇒ c (water) = 4.184 J/g°C

    ⇒ΔT (water) = T2 - 20.3 °C

    1100 * 0.803 * (T2 - 87.3) = - 2000 * 4.184 * (T2 - 20.3)

    883.3 (T2 - 87.3) = - 8368 (T2 - 20.3)

    883.3T2 - 77112.09 = - 8368T2 + 169870.4

    883.3T2 + 8368T2 = 169870.4 + 77112.09

    9251.3T2 = 246982.49

    T2 = 26.7 °C

    The final temperature will be 26.7 °C
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