THERMAL CYU 4

Physical Situation: We are dealing with a moving boundary problem where a liquid freezes progressively from left to right. The latent heat released during freezing is the source of heat that raises the temperature of the entire system (both the frozen and liquid parts).

Let the cross-sectional area of the box be $A = b \times h$. The density of the substance is $\rho$ in both states. At any time $t$, the boundary has moved a distance $x = vt$.

• Mass of frozen part: $m_s = \rho A (vt)$
• Mass of liquid part: $m_l = \rho A (l – vt)$

Specific Heat Capacities:

• Liquid specific heat: $s$
• Solid specific heat is $\eta = 10\%$ less than liquid. This means the specific heat of the solid is $s(1 – \eta)$.

Energy Balance Equation:

In a small time interval $dt$, a small mass $dm$ freezes. The heat released by this freezing is used to raise the temperature of the entire mass (solid + liquid) by $d\theta$.

$$ \text{Heat Released} = \text{Heat Absorbed} $$

$$ L \cdot dm = (m_s s_{solid} + m_l s_{liquid}) d\theta $$

Substituting the expressions for mass and specific heat:

$$ L (\rho A v dt) = [\rho A vt \cdot s(1-\eta) + \rho A (l-vt) \cdot s] d\theta $$

Canceling $\rho A$ from both sides and simplifying the term in the brackets:

$$ vL dt = s [ vt(1-\eta) + (l-vt) ] d\theta $$

$$ vL dt = s [ vt – \eta vt + l – vt ] d\theta $$

$$ vL dt = s [ l – \eta vt ] d\theta $$

Now, we rearrange to integrate. We separate the variables $\theta$ and $t$:

$$ d\theta = \frac{vL}{s(l – \eta vt)} dt $$

Integration:

Integrate from initial state ($t=0, \theta=\theta_0$) to final state ($t, \theta$):

$$ \int_{\theta_0}^{\theta} d\theta = \frac{vL}{s} \int_{0}^{t} \frac{dt}{l – \eta vt} $$

To integrate the right side, recall $\int \frac{dx}{a – bx} = -\frac{1}{b} \ln(a-bx)$. Here, the variable is $t$ and coefficient is $-\eta v$.

$$ \theta – \theta_0 = \frac{vL}{s} \left[ \frac{\ln(l – \eta vt)}{-\eta v} \right]_{0}^{t} $$

$$ \theta – \theta_0 = -\frac{L}{\eta s} \left[ \ln(l – \eta vt) – \ln(l) \right] $$

Using property $\ln A – \ln B = \ln(A/B)$:

$$ \theta = \theta_0 – \frac{L}{\eta s} \ln\left( \frac{l – \eta vt}{l} \right) $$