Find the change in the internal energy of 2kg of water as it is heated

1.	Find the change in the internal energy of 2kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200JAg-K and its densities at 0°C and 4°C are 999.9kg/m3 and 1000kg/m3 respectively. Atmospheric pressure = 105 Pa.
Answer» Given M = 2kg 2t = 4°c Sw = 4200J/Kg–k f₀= 999.9kg/m³ f₄ = 1000kg/m³ P = 10⁵Pa. Net internal energy = dv dQ = DU + dw => msΔQϕ = dU + P(v₀ – v₄) => 33600 = du + 10⁵(m/v₀ - m/v₄) = du + (0.0020002 - 0.002) = du + 10⁵0.0000002 => 2 × 4200 × 4 = dU + 10⁵(m – m) => 33600 = du + 0.02 => du = (33600 - 0.02)J

Find the change in the internal energy of 2kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200JAg-K and its densities at 0°C and 4°C are 999.9kg/m3 and 1000kg/m3 respectively. Atmospheric pressure = 105 Pa.

Answer»

Given M = 2kg

2t = 4°c

Sw = 4200J/Kg–k

f₀= 999.9kg/m³

f₄ = 1000kg/m³

P = 10⁵Pa.
Net internal energy = dv
dQ = DU + dw => msΔQϕ = dU + P(v₀ – v₄)

=> 33600 = du + 10⁵(m/v₀ - m/v₄) = du + (0.0020002 - 0.002) = du + 10⁵0.0000002
=> 2 × 4200 × 4 = dU + 10⁵(m – m)

=> 33600 = du + 0.02 => du = (33600 - 0.02)J