A student removed a 49.21 g piece of aluminum (Cp = 0.897 J/(g°C)) from a boilin

Chemistry
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A student removed a 49.21 g piece of aluminum (Cp = 0.897 J/(g°C)) from a boiling water bath and placed it into a styrofoam cup containing 90. g of water at 22.3°C. Determine the equilibrium temperature (to 0.1°C) of the water and aluminum.

Mar 18th, 2015

The heat lost by the aluminum as it cools must be equal to the heat acquired by water. Let the equilibrium temperature be t. Then C_p, Al * m_Al * (100 - t) = C_P, H2O * m_H2O * t, where C_P, H2O = 4.187 J/(g*C), and m stands for masses. From the above equation,

t = C_P, Al * m_Al * 100 / (C_P, Al * m_Al + C_P, H2O * m_H2O ) = 0.897 * 49.21 *100 / (0.897 * 49.21 + 4.187 * 90) = 10.5 °C

Mar 17th, 2015

Need to correct my answer (the initial temperature of water was not taken into account).

The heat lost by the aluminum as it cools must be equal to the heat acquired by water. Let the equilibrium temperature be t. Then C_p, Al * m_Al * (100 - t) = C_P, H2O * m_H2O *( t - 22.3), where C_P, H2O = 4.187 J/(g*C), and m stands for masses. From the above equation,

t = (C_P, Al * m_Al * 100 + C_P, H2O * m_H2O * 22.3) / (C_P, Al * m_Al + C_P, H2O * m_H2O ) =

(0.897 * 49.21 *100 + 4.187 * 90 * 22.3) / (0.897 * 49.21 + 4.187 * 90) = 30.4 °C



Mar 17th, 2015

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