NEET-XII-Physics
26: Laws of Thermodynamics
- #19Find 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.Ans : Given:
Mass of water, M = 2 kg
Change in temperature of the system, ∆θ = 4°C = 277 K
Specific heat of water, sw = 4200 J/kg-°C
Initial density, p0 = 999.9 kg/m3
Final density, pf = 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
Page No 63: