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Standard XII

Physics

Area and Volume Expansion

A metal of ma...

Question

A metal of mass 1 kg at constant atmospheric pressure and at initial temperature $20^{o} C$ is given the heat of $20, 000 J$ . Find the Change in internal energy.
Given: Specific heat $400$ $J / k g /^{o} C$ .
Coefficient of cubical expansion $γ = 9 \times 10^{- 5} /^{o} C$
Density $p = 9000 k g / m^{3}$ , atmospheric pressure $= 10^{5} N / m^{2}$

19, 999.95 J

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18, 999.95 J

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20, 999.95 J

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None of these

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Solution

The correct option is B $19, 999.95 J$
20000= $m c Δ T \Rightarrow Δ T = 50^{o} C$
so $T_{f} = 20 + 50 = 70^{o} C$
Now change in volume due to heating $Δ V = V γ δ T = \frac{m γ Δ T}{ρ} = 5 \times 10^{- 7} m^{3}$
Work done= $P Δ V = 0.05 J$
Now $Δ = Q - W = 20000 - 0.05 = 19999.95 J$

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A metal of mass 1 kg at constant atmospheric pressure and at initial temperature 20oC is given the heat of 20,000J. Find the Change in internal energy.Given: Specific heat 400J/kg/oC.Coefficient of cubical expansion γ=9×10−5/oCDensity p=9000kg/m3, atmospheric pressure =105N/m2

The correct option is B 19,999.95J20000=mcΔT⇒ΔT=50oCso Tf=20+50=70oCNow change in volume due to heating ΔV=VγδT=mγΔTρ=5×10−7m3Work done=PΔV=0.05JNow Δ=Q−W=20000−0.05=19999.95J

The correct option is B $19, 999.95 J$
20000= $m c Δ T \Rightarrow Δ T = 50^{o} C$
so $T_{f} = 20 + 50 = 70^{o} C$
Now change in volume due to heating $Δ V = V γ δ T = \frac{m γ Δ T}{ρ} = 5 \times 10^{- 7} m^{3}$
Work done= $P Δ V = 0.05 J$
Now $Δ = Q - W = 20000 - 0.05 = 19999.95 J$