A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. When the flas

Question

A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0°C , a volume of 8.50 cm^3 of mercury overflows the flask.Required:If the coefficient of volume expansion of mercury is βHg = 1.80×10^−4 /K , compute βglass, the coefficient of volume expansion of the glass. Express your answer in inverse kelvins.

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Eirian 4 years 2021-08-22T09:17:57+00:00 1 Answers 24 views 0

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  1. Helga
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    2021-08-22T09:18:59+00:00
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    Answer:

    the coefficient of volume expansion of the glass is \mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}

    Explanation:

    Given that:

    Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

    temperature of the glass flask and mercury= 1.00° C

    After heat is applied ; the final temperature = 52.00° C

    Temperature change ΔT = 52.00° C – 1.00° C = 51.00° C

    Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

    the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

    The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

    The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

    Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

    Now; the mercury overflow = Increase in volume of the mercury – increase in the volume of the flask

    the mercury overflow = (9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3

    8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3

    8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

    -6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

    6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

    \dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )

    \mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}

    Thus; the coefficient of volume expansion of the glass is \mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}

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