A cube of side 5 cm made of iron and having a mass of 1500 g is heated from 25° C to 400° C. The specific heat for iron is 0.12 cal / g ° C and the coefficient of volume expansion is 3.5 × 10-5 / ° C, the change in the internal energy of the cube is ( atm. pressure =1 × 105 N / m 2 )

Question

A cube of side 5 cm made of iron and having a mass of 1500 g is heated from 25° C to 400° C. The specific heat for iron is 0.12 cal / g ° C and the coefficient of volume expansion is 3.5 × 10-5 / ° C, the change in the internal energy of the cube is ( atm. pressure =1 × 105 N / m 2 ) (A) 320 kJ (B) 282 kJ (C) 141 kJ (D) 4023 kJ

Tardigrade Admin · Accepted Answer

Q =m C Δ T=1.5 × 0.12 × 4200 ×(400-25) =2.83 × 105 J W =P(Δ V)=P(V &gamma; Δ T) =105 ×(5 × 10-2)3 × 3.5 × 10-5 × 375=0.164 J Thus Q=Δ U+W or 2.83 × 105=Δ U+0.164 ; Δ U =282 kJ

$320 k J$

$282 k J$

$141 k J$

$4023 k J$

$Q = m C Δ T = 1.5 \times 0.12 \times 4200 \times (400 - 25)$
$= 2.83 \times 1 0^{5} J$
$W = P (Δ V) = P (Vγ Δ T)$
$= 1 0^{5} \times (5 \times 1 0^{- 2})^{3} \times 3.5 \times 1 0^{- 5} \times 375 = 0.164 J$
Thus $Q = Δ U + W$
or $2.83 \times 1 0^{5} = Δ U + 0.164; Δ U = 282 k J$

Q. A cube of side 5cm made of iron and having a mass of 1500g is heated from 25∘C to 400∘C. The specific heat for iron is 0.12cal/g∘C and the coefficient of volume expansion is 3.5×10−5/∘C, the change in the internal energy of the cube is (atm pressure =1×105N/m2 )

320kJ

282kJ

141kJ

4023kJ

Q=mCΔT=1.5×0.12×4200×(400−25) =2.83×105J W=P(ΔV)=P(VγΔT) =105×(5×10−2)3×3.5×10−5×375=0.164J Thus Q=ΔU+W or 2.83×105=ΔU+0.164;ΔU=282kJ

$320 k J$

$282 k J$

$141 k J$

$4023 k J$

$Q = m C Δ T = 1.5 \times 0.12 \times 4200 \times (400 - 25)$
$= 2.83 \times 1 0^{5} J$
$W = P (Δ V) = P (Vγ Δ T)$
$= 1 0^{5} \times (5 \times 1 0^{- 2})^{3} \times 3.5 \times 1 0^{- 5} \times 375 = 0.164 J$
Thus $Q = Δ U + W$
or $2.83 \times 1 0^{5} = Δ U + 0.164; Δ U = 282 k J$