Solution: Calorimetry and Heating Curves
1. Mass Distribution Analysis
We are given a total mass $m = 0.60 \text{ kg}$ consisting of crushed ice and supercooled water. The problem states that the heat capacities of the water and ice components in the calorimeter are equal.
$$ m_i s_i = m_w s_w $$Using the specific heat values $s_i = 2.1 \text{ kJ/(kg}\cdot\text{K)}$ and $s_w = 4.2 \text{ kJ/(kg}\cdot\text{K)}$:
$$ s_w = 2 s_i $$ $$ m_i s_i = m_w (2s_i) \implies m_i = 2m_w $$Substituting this into the total mass equation:
$$ m_i + m_w = 0.60 \text{ kg} $$ $$ 2m_w + m_w = 0.60 \implies 3m_w = 0.60 $$ $$ m_w = 0.20 \text{ kg}, \quad m_i = 0.40 \text{ kg} $$2. Calculation of Heater Power ($P$)
From the graph, the temperature of the mixture rises from $-2^{\circ}\text{C}$ to $0^{\circ}\text{C}$ in the time interval $t = 0$ to $t = 20 \text{ s}$.
$$ \frac{d\theta}{dt} = \frac{\Delta \theta}{\Delta t} = \frac{0 – (-2)}{20} = \frac{2}{20} = 0.1 \text{ }^{\circ}\text{C/s} $$The power supplied by the heater raises the temperature of both the ice and the water:
$$ P = (m_i s_i + m_w s_w) \frac{d\theta}{dt} $$Calculating the total heat capacity ($C_{total}$):
$$ C_{total} = (0.40 \times 2100) + (0.20 \times 4200) $$ $$ C_{total} = 840 + 840 = 1680 \text{ J/}^{\circ}\text{C} $$Now, calculate Power:
$$ P = 1680 \times 0.1 = 168 \text{ W} $$3. Time to Melt Ice ($t_1$)
At $0^{\circ}\text{C}$, the supplied heat is used to melt the ice (latent heat). The temperature remains constant during this phase.
$$ P \cdot t_1 = m_i L $$Given $L = 0.33 \text{ MJ/kg} = 330,000 \text{ J/kg}$:
$$ 168 \cdot t_1 = 0.40 \times 330,000 $$ $$ t_1 = \frac{132,000}{168} = 785.71 \text{ s} $$4. Time to Heat Water to $20^{\circ}\text{C}$ ($t_2$)
After the ice melts completely, we have $0.60 \text{ kg}$ of water. We need to find the time $t_2$ to raise the temperature from $0^{\circ}\text{C}$ to $20^{\circ}\text{C}$.
$$ P \cdot t_2 = m_{total} s_w \Delta \theta $$ $$ 168 \cdot t_2 = 0.60 \times 4200 \times (20 – 0) $$ $$ 168 \cdot t_2 = 50,400 $$ $$ t_2 = \frac{50,400}{168} = 300 \text{ s} $$Final Results:
- Power of Heater ($P$): $168 \text{ W}$
- Time to melt ice ($t_1$): $785.71 \text{ s}$
- Time to reach $20^{\circ}\text{C}$ ($t_2$): $300 \text{ s}$