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A cube of side 5 cm made of iron and having a mass of 1500 gm, is heated from $250^{0}$ C to 400^oC. The specific heat for iron is 0.12 cal/gm/°C and the coefficient of volume expansion is 3.5×10^–5/°C. The change in internal energy of the cube is: (atmospheric pressure = 10⁵ N/m²)

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320 kJ

282 kJ

141 kJ

423 kJ

answer is B.

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Detailed Solution

1 cal/gm = 4200 J/kg.

Q = 1500 \times {10^{ - 3}} \times 0.12 \times 4200 \times (400 - 25)

= 15×12×42×375× ${10^{- 1}}^{}$ = 283500 J

Q = 283.5 kJ

$∆ W = P ∆ V = 10^{5} V γ ∆ t = 10^{5} \times 125 \times 10^{- 6} \times 3.5 \times 10^{- 5} \times 375 = 0.164 J$ it is very very small compare to Q

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