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