Thermodynamics Question 238

Question: A cube of side 5 cm made of iron and having a mass of 1500 g is heated from $ 25{}^\circ C $ to $ 400{}^\circ C. $ The specific heat for iron is $ 0.12cal/g{}^\circ C $ and the coefficient of volume expansion is $ 3.5\times {{10}^{-5}}/{}^\circ C, $ the change in the internal energy of the cube is (atm pressure $ 1\times 10^{5}N/m^{2} $ )

Options:

A) 320 kJ

B) 282 kJ

C) 141 kJ

D) 423 kJ

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $ Q=mC\Delta r=1.5\times 0.12\times 4200\times ( 400-25 ) $

$ =2.83\times 10^{5}J $

$ W=P( \Delta V )=P(V\Delta T) $

$ =105\times ( 5\times 10-2 )3\times 3.5\times 10-5\times 375=0.164J $

$ \text{Thus }Q=\Delta U+W $

$ or2.83\times 10^{5}=\Delta U+0.164;\Delta U=282kJ $



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