Question

Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ and its densities at 0°C and 4°C are 999.9 kg m⁻³ and 1000 kg m⁻³ respectively. Atmospheric pressure = 10⁵ Pa.

Answer 1

Given:-

Mass of water, M = 2 kg

Change in temperature of the system, ∆θ = 4°C = 277 K

Specific heat of water, s_w = 4200 J/kg-°C

Initial density, p₀ = 999.9 kg/m³

Final density, p_f = 1000 kg/m³

P = 10⁵ Pa

Let change in internal energy be ∆U.

Using the first law of thermodynamics, we get

∆Q = ∆U + ∆W

Also, ∆Q = ms∆θ

W = P∆V = P(V_f - V_i)

⇒ ms∆θ = ∆U + P (V₀ − V₄)

⇒ 2 × 4200 × 4= ∆U + 10⁵ (∆V)

⇒ 33600 = ∆U + 10⁵

\[\left( \frac{m}{p_0} - \frac{m}{p_f} \right)\]

⇒ 33600 = ∆U + 10⁵ × ( - 0.0000002)

⇒ 33600 = ∆U - 0.02

∆U = (33600 - 0.02) J