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:
$ Q=mC\Delta T=1.5\times 0.12\times 4200\times ( 400-25 ) $
$ =2.83\times 10^{5}J $
$ W=P( \Delta V )=P(\Delta V \Delta T) $
$ =105\times ( 5\times 10-2 ) \times 3\times 3.5\times 10^{-5}\times 375=0.164J $
$ \text{Thus }Q=\Delta U+W $
$ or2.83\times 10^{5}=\Delta U+164;\Delta U=282kJ $
BETA
