1 kg of water at 10 0 ∘ C ( V 1 = 1.0 × 1 0 − 3 m 3 ) turns completely into steam ( V 2 = 1.673 m 3 ) at 1.013 × 1 0 5 Pa. The latent heat is 2.26 × 1 0 6 J/kg. Find the change in internal energy in units of 1 0 6 J (two decimals).

Δ U = L − P 0 ( V 2 − V 1 ) = 2.26 × 1 0 6 − 1.013 × 1 0 5 × 1.672 = 2.09 × 1 0 6 J.

Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesMedium2 minQB

PhysicsMediumnumerical

1 kg of water at $10 0^{\circ}$ C ( $V_{1} = 1.0 \times 1 0^{- 3}$ $m^{3}$ ) turns completely into steam ( $V_{2} = 1.673$ $m^{3}$ ) at $1.013 \times 1 0^{5}$ Pa. The latent heat is $2.26 \times 1 0^{6}$ J/kg. Find the change in internal energy in units of $1 0^{6}$ J (two decimals).

Answer:

2.09

Solution:

$Δ U = L - P_{0} (V_{2} - V_{1}) = 2.26 \times 1 0^{6} - 1.013 \times 1 0^{5} \times 1.672 = 2.09 \times 1 0^{6}$ J.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics Laws & Specific Heats of Gases

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...