A piece of aluminum metal at an initial temperature of 97.3°C was placed in a calorimeter containing 49.0g of water at an initial temperatur

Question

A piece of aluminum metal at an initial temperature of 97.3°C was placed in a calorimeter containing 49.0g of water at an initial temperature of 22.0°C. The two were allowed to come to thermal equilibrium and the final temperature was 24.1°C. The specific heats of water and aluminum are 4.184 J/g °C and 0.902 J/g °C, respectively. What was the mass of the Al piece that was added?

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Edana Edana 5 years 2021-08-29T17:08:00+00:00 1 Answers 15 views 0

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  1. thnahdat
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    2021-08-29T17:09:22+00:00
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    Answer:

    0.153 g

    Explanation:

    Because the aluminum metal was at a higher temperature, it loses heat while the water gains heat. By the principle of mixtures, heat lost by hot material is the same heat gained by colder material, assuming there is no heat loss to the environment.

    We work under the assumption that the heat gained by the calorimeter is negligible.

    Heat lost by metal

    H_\text{Al} = m_\text{Al}\times c_\text{Al}\times\Delta\theta_\text{Al}

    H_\text{Al} = m_\text{Al}\times 0.902\times(97.3-24.1) = 66.0264m_\text{Al}

    where m, c and Δθ are the mass, specific heat  and change in temperature of the aluminum.

    Heat gained by water

    H_\text{W} = m_\text{W}\times c_\text{W}\times\Delta\theta_\text{W}

    H_\text{W} = 49.0\times 4.184\times(24.1 - 22.0) = 430.5336\text{ J}

    Equating both energy equations,

    66.0264m_\text{Al} = 430.5336

    m_\text{Al} = 0.153\text{ g}

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