26.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 21.0 ∘C in an insulated cup.What will

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26.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 21.0 ∘C in an insulated cup.What will the new water temperature be?

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Thu Giang 6 months 2021-07-28T04:55:00+00:00 1 Answers 12 views 0

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    2021-07-28T04:56:13+00:00

    Answer:

    The new water temperature is 26.4 °C

    Explanation:

    Given;

    mass of copper, M_{cu} = 26 g = 0.026 kg

    temperature of copper, t = 300 °C

    volume of water, V = 120 mL = 0.12 L

    temperature of water, t = 21 °C

    density of water, ρ = 1 kg/L

    mass of water = density x volume

    mass of water = (1 kg/L) x 0.12 L = 0.12 kg

    heat lost by copper = heat gained by water

    Both copper and water reach final temperature, T

    Heat gained by water, Q_w = m_wcΔθ = m_w C(T - t)

    Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)

    Heat lost by copper is given by;

    Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)

    Q_{cu} = Q_w

    504(T- 21) = 10.01(300 – T)

    504 T – 10584 = 3003 – 10.01 T

    504 T + 10.01 T= 3003 + 10584

    514.01 T = 13587

    T = (13587) / 514.01

    T = 26.4 °C

    Therefore, the new water temperature is 26.4 °C

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