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Question Number 27769 by tawa tawa last updated on 14/Jan/18

$$\mathrm{A}\mathrm{glass}\mathrm{bottle}\mathrm{full}\mathrm{of}\mathrm{mercury}\mathrm{has}\mathrm{mass}500\mathrm{g}.\mathrm{On}\mathrm{being}\mathrm{heated}\mathrm{through}35°\mathrm{C},$$$$2.43\mathrm{g}\mathrm{of}\mathrm{mercury}\mathrm{are}\mathrm{expelled}.\mathrm{calculate}\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{mercury}\mathrm{remaining}\mathrm{in}$$$$\mathrm{the}\mathrm{bottle}(\mathrm{Cubic}\mathrm{expansivity}\mathrm{of}\mathrm{mercury}\mathrm{is}1.8\times {10}^{-4}\mathrm{per}\mathrm{K}.$$$$\mathrm{linear}\mathrm{expansivity}\mathrm{of}\mathrm{glass}\mathrm{is}8.0\times {10}^{-6}\mathrm{per}\mathrm{K}.$$

Answered by mrW2 last updated on 15/Jan/18

$$V=volumeofmercuryinglassbeforeheating$$$$\mathrm{\Delta }V=changeofvolumeofmercuryduetoheating$$$$\mathrm{\Delta }V/V=(1.8\times {10}^{-4}-3\times 8.0\times {10}^{-6})\times 35=0.00546$$$$\frac{V}{\mathrm{\Delta }V}=\frac{1}{0.00546}=183.15$$$$Massofmercuryinglassbeforeheating$$$$M=183.15\times 2.43=445.05g$$$$Massofmercuryremaininginglassafterheating$$$$M^{\prime }=445.05-2.43=442.62g$$

Commented by tawa tawa last updated on 15/Jan/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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