THERMAL BYU 3

1. Initial Analysis & Data Extraction

We are given a mixture of equal masses $m$ of ice ($I$) and an unknown substance ($S$).
Initial temperature: $-40^\circ\text{C}$.
Specific heats in solid state: $s_i = 2100 \text{ J}/(\text{kg}\cdot^\circ\text{C})$, $s_s = 900 \text{ J}/(\text{kg}\cdot^\circ\text{C})$.

From the Temperature-Time graph, we observe three distinct phases:

1. $0$ to $t_1=1$ min: The temperature rises from $-40^\circ\text{C}$ to $-20^\circ\text{C}$. Both ice and substance are solids.
2. $t_1$ to $t_2$: The temperature remains constant at $-20^\circ\text{C}$. Since pure ice melts at $0^\circ\text{C}$, this plateau corresponds to the melting of the unknown substance.
3. $t_2$ to $t_3=4$ min: The temperature rises from $-20^\circ\text{C}$ to $0^\circ\text{C}$. The mixture contains solid ice and liquid substance.

Visually inspecting the time axis grids, the plateau ends at approximately $t_2 \approx 2.67 \text{ min}$ (or $8/3$ min). We will verify this consistency in the calculation.

2. Phase 1: Heating Solids ($-40^\circ\text{C} \to -20^\circ\text{C}$)

Let $P$ be the constant rate of heat supply (Power). Total heat capacity of the system is $m(s_i + s_s)$. Duration $\Delta t = 1 \text{ min}$. $\Delta \theta = 20^\circ\text{C}$.

$$ P \cdot (1) = m(s_i + s_s) \Delta \theta $$ $$ \frac{P}{m} = (2100 + 900)(20) = 3000 \times 20 = 60,000 \text{ J}/(\text{kg}\cdot\text{min}) $$

3. Phase 2: Substance Melting (at $-20^\circ\text{C}$)

Let $L_f$ be the latent heat of fusion of the substance. Duration is $(t_2 – t_1)$. Using the graph reading $t_2 \approx 8/3$ min: $$ \Delta t_{melt} = \frac{8}{3} – 1 = \frac{5}{3} \text{ min} $$

Energy balance:

$$ P \cdot \left(\frac{5}{3}\right) = m L_f $$ $$ L_f = \frac{P}{m} \times \frac{5}{3} $$ Substituting $P/m = 60,000$: $$ L_f = 60,000 \times 1.67 \approx 100,000 \text{ J}/\text{kg} $$

4. Phase 3: Heating Liquid Substance + Solid Ice ($-20^\circ\text{C} \to 0^\circ\text{C}$)

Here, ice is still solid ($s_i = 2100$), but the substance is liquid with unknown specific heat $s_{s(\text{liq})}$. Duration $\Delta t = 4 – 8/3 = 4/3 \text{ min}$. $\Delta \theta = 20^\circ\text{C}$.

$$ P \cdot \left(\frac{4}{3}\right) = m(s_i + s_{s(\text{liq})}) (20) $$ $$ \frac{P}{m} \cdot \frac{4}{3} = 20 (2100 + s_{s(\text{liq})}) $$ $$ 60,000 \cdot \frac{4}{3} = 20 (2100 + s_{s(\text{liq})}) $$ $$ 80,000 = 20 (2100 + s_{s(\text{liq})}) $$ $$ 4000 = 2100 + s_{s(\text{liq})}$$ $$ s_{s(\text{liq})} = 1900 \text{ J}/(\text{kg}\cdot^\circ\text{C}) $$