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9 July, 13:42

1.2 kg of aluminum at 20oC is added to 1.5 kg of water at 80oC. After the system reaches thermal equilibrium, what is its final temperature (in oC) ?

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  1. 9 July, 15:56
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    The final temperature is 71.19 °C

    Explanation:

    Step 1: Data given

    Mass of aluminium = 1.2 kg = 1200 grams

    Temperature of aluminium = 20.0 °C

    Specific heat of aluminium = 0.900 J/g°C

    Mass of water = 1.5 kg = 1500 grams

    Temperature of water = 80.0 °C

    Specific heat of water = 4.184 J/g°C

    Step 2: Calculate the final temperature

    heat gained = heat lost

    Q (aluminium) = - Q (water)

    Q = m*c*ΔT

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

    ⇒ with mass of aluminium = 1200 grams

    ⇒ with specific heat of aluminium = 0.900 J/g°C

    ⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 20.0 °C

    ⇒ with mass of water = 1500 grams

    ⇒ with specific heat of water = 4.184 J/g°C

    ⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 80.0°C

    1200 * 0.900 * (T2-20.0°C) = - 1500 * 4.184 * (T2 - 80.0°C)

    1080 * (T2 - 20.0°C) = - 6276 * (T2 - 80.0°C)

    1080 T2 - 21600 = - 6276T2 + 502080

    7356T2 = 523680

    T2 = 71.19 °C

    The final temperature is 71.19 °C
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