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)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

320 kJ

282 kJ

141 kJ

423 kJ

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

1 cal/gm = 4200 J/kg.

= 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

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE