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

The change in the internal energy is Δu = (33600 - 0.02)J

Explanation:

Given M = 2kg  

2t = 4°c  

Sw = 4200J/Kg–k

f0 = 999.9kg/m3  

f4 = 1000kg/m3  

P = 105Pa

Net internal energy = Δv

ΔQ = ΔU + vw => msΔQθ = vU + P(v0 – v4)

33600 = du + 10^5(m/v0 - m/v4)

= Δu + (0.0020002 - 0.002)

= Δu + 10^50.0000002

2 × 4200 × 4 = ΔU + 10^5(m – m)

33600 = Δu + 0.02

Δu = (33600 - 0.02)J

Thus the change in the internal energy is Δu = (33600 - 0.02)J

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Answer 2

Explanation:

Step 1:

M = 2 kg

∆θ = 4°C = 277 K

$\begin{aligned}&s_{w}=4200 \mathrm{J} / \mathrm{kg} \^ - ^{\circ} \mathrm{C}\\&p_{0}=999.9 \mathrm{kg} / \mathrm{m}^{3}\\&p_{f}=1000 \mathrm{kg} / \mathrm{m}^{3}\end{aligned}$

M  is Mass of water,  

∆θ   is Change in temperature of the system

S$_w$ is Specific heat of water,  

P$_0$ is  Initial density,  

$p_f$ Is Final density,  

Step 2:

$P=10^{5} \mathrm{Pa}$

Let ∆U is  change in internal energy.

Step 3:

Using the thermodynamics first law, we get

∆Q = ∆U + ∆W

∆Q = ms∆θ

$\begin{aligned}&\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{P}\left(V_{f}-V_{i}\right)\\&\mathrm{ms} \Delta \theta=\Delta \mathrm{U}+\mathrm{P}\left(V_{0}-V_{4}\right)\\&2 \times 4200 \times 4=\Delta U+10^{5}(\Delta V)\\& 33600=\Delta \mathrm{U}+10^{5}\left(\frac{m}{p_{0}}-\frac{m}{p_{f}}\right)\\&33600=\Delta U+10^{5} \times(-0.0000002)\end{aligned}$

$\begin{aligned}&33600=\Delta \mathrm{U}-0.02 |\\&\Delta \mathrm{U}=(33600-0.02) J\end{aligned}$