THERMAL BYU 9

We are given a mixture of water and ice heated by a constant power source.

• Power of heater, $P = 50 \text{ W}$.
• Specific heat of water, $s_w = 4.0 \text{ J}/(\text{g}^\circ\text{C})$.
• Latent heat of fusion of ice, $L_f = 390 \text{ J/g}$.

• $t = 2 \text{ min} \rightarrow T = 0^\circ\text{C}$
• $t = 3 \text{ min} \rightarrow T = 2^\circ\text{C}$
• $t = 4 \text{ min} \rightarrow T = 7^\circ\text{C}$

Between $t = 3 \text{ min}$ and $t = 4 \text{ min}$, the temperature rises from $2^\circ\text{C}$ to $7^\circ\text{C}$. In this interval, all ice has melted, and we are heating the total mass of water ($M_{total}$).

$$ \Delta t = 60 \text{ s}, \quad \Delta T = 7 – 2 = 5^\circ\text{C} $$

Energy supplied ($E$) equals heat absorbed ($Q$):

$$ E = P \times \Delta t = 50 \times 60 = 3000 \text{ J} $$

Using $Q = ms\Delta T$:

$$ 3000 = M_{total} \times 4.0 \times 5 $$

$$ 3000 = 20 M_{total} \implies M_{total} = 150 \text{ g} $$

Between $t = 2 \text{ min}$ and $t = 3 \text{ min}$, the temperature rises from $0^\circ\text{C}$ to $2^\circ\text{C}$.

Energy supplied in this interval is also $3000 \text{ J}$ (since $\Delta t = 60 \text{ s}$).

Energy required to raise the temperature of $150 \text{ g}$ of water from $0^\circ\text{C}$ to $2^\circ\text{C}$:

$$ Q_{heat} = 150 \times 4.0 \times (2 – 0) = 1200 \text{ J} $$

The excess energy supplied must have been utilized to melt the remaining portion of ice ($m_{rem}$) that existed at $t=2 \text{ min}$:

$$ E_{melting} = 3000 – 1200 = 1800 \text{ J} $$

$$ m_{rem} \times L_f = 1800 \implies m_{rem} = \frac{1800}{390} = \frac{60}{13} \text{ g} $$

In the first 2 minutes ($t=0$ to $t=120 \text{ s}$), the temperature remained at $0^\circ\text{C}$. The energy supplied was used purely for melting ice.

$$ E_{initial} = 50 \times 120 = 6000 \text{ J} $$

Mass of ice melted in first 2 mins ($m_{melted\_1}$):

$$ m_{melted\_1} = \frac{6000}{390} = \frac{200}{13} \text{ g} $$

Total initial mass of ice ($m_{ice}$) is the sum of ice melted initially and the ice remaining at 2 mins:

$$ m_{ice} = m_{melted\_1} + m_{rem} = \frac{200}{13} + \frac{60}{13} = \frac{260}{13} = 20 \text{ g} $$