Determine the change in the internal energy of 2 kg of water when it is heated from 0^ C to 4^ C . The specific heat capacity of water is 4200 J kg ⁻¹ K ⁻¹ , while its densities at 0^ C and 4^ C are 999.9 kg m ⁻³ and 1000 kg m ⁻³ , respectively. The atmospheric pressure is 10^5 Pa .

Step-by-step solution

The heat supplied to the water is given by Q = m c T . Substituting the given values: Q = 2 4200 (4 - 0) = 33600 J The change in volume of the water is V = V₂ - V₁ = m ( 1 ₂ - 1 ₁ ) . Substituting the densities: V = 2 ( 1 1000 - 1 999.9 ) = 2 ( 999.9 - 1000 1000 999.9 ) = -0.2 999900 -2 10⁻⁷ m ^3 The work done by the water against the atmospheric pressure is W = P V . W = 10^5 (-2 10⁻⁷) = -0.02 J According to the first law of thermodynamics, Q = U + W . U = Q - W = 33600 - (-0.02) = 33600.02 J