When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coefficient of volu

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When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coefficient of volume expansion? PLEASE HELPPP MEEEEE

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Khánh Gia 5 years 2021-09-04T16:18:08+00:00 1 Answers 32 views 0

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  1. Dulcie
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    2021-09-04T16:19:30+00:00
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    Coefficient of volume expansion is 8.1 ×10⁻⁴ C⁻¹.

    Explanation:

    The volume expansion of a liquid is given by ΔV,

    ΔV = α V₀ ΔT

    ΔT = change in temperature  = 48.5° C

    α =  coefficient of volume expansion =?

    V₀ = initial volume = 2.35 m³

    We need to find α , by plugin the given values in the equation and by rearranging the equation as,

    \alpha=\frac{\Delta \mathrm{V}}{\mathrm{V}_{0} \Delta \mathrm{T}}=\frac{0.0920}{2.35 \times 48.5}=0.00081

      = 8.1 ×10⁻⁴ C⁻¹.

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