t=t1+t2=MR(CΔT+L) A glass flask is filled up to a mark with 50 cc of mercury at 18∘C. If the flask and contents are heated to 38∘C, how much mercury will be above the mark? (α for glass is 9×10−6/∘C and coefficient of real expansion of mercury is180×10−6/∘C)

0.85 cc

0.46 cc

0.153 cc

0.05 cc

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t=t1+t2=MR(CΔT+L) A glass flask is filled up to a mark with 50 cc of mercury at 18∘C. If the flask and contents are heated to 38∘C, how much mercury will be above the mark? (α for glass is 9×10−6/∘C and coefficient of real expansion of mercury is180×10−6/∘C)

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Text solutionVerified

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by

Δ V = V 0 [γ L - γ] Δ θ = V [γ m - 3 α g] Δ θ

=50[180×10−6−3×9×10−6](38−18)

= 0.153

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Question Text	t=t1+t2=MR(CΔT+L) A glass flask is filled up to a mark with 50 cc of mercury at 18∘C. If the flask and contents are heated to 38∘C, how much mercury will be above the mark? (α for glass is 9×10−6/∘C and coefficient of real expansion of mercury is180×10−6/∘C)
Answer Type	Text solution:1
Upvotes	150

t=t1+t2=MR(CΔT+L) A glass flask is filled up to a mark with 50 cc of mercury at 18∘C. If the flask and contents are heated to 38∘C, how much mercury will be above the mark? (α for glass is 9×10−6/∘C and coefficient of real expansion of mercury is180×10−6/∘C)

Text solutionVerified

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