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26 August, 11:22

A 155.0 g piece of copper at 179 oC is dropped into 250.0 g of water at 28.1 oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.)

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  1. 26 August, 11:51
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    The final temperature is 36.24 °C

    Explanation:

    Step 1: Data given

    Mass of copper = 155.0 grams

    Initial temperature of copper = 179.0 °C

    Mass of water = 250.0 grams

    The initial temperature of water = 28.1 °C

    The specific heat of copper = 0.385 J/g°C

    The specific heat of water = 4.184 J/g°C

    Step 2: Calculate the final temperature

    Qlost = - Qgained

    Q = m*c*ΔT

    Qcopper = - Qwater

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

    ⇒with mass of copper = 155.0 grams

    ⇒with c (copper) = the specific heat of copper = 0.385 J/g°C

    ⇒with ΔT = the change of temperature of copper = T2 - T1 = T2 - 179.0°C

    ⇒with mass water = 250.0 grams

    ⇒with c (water) = the specific heat of water = 4.184 J/g°C

    ⇒with ΔT = with ΔT = the change of temperature of copper = T2 - T1 = T2 - 28.1 °C

    155.0 * 0.385 * (T2 - 179.0°C) = - 250 * 4.184 * (T2 - 28.1 °C)

    T2 = 36.24 °C

    The final temperature is 36.24 °C
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