1. How much energy is associated with 95 g of ice at $$ -12^{\circ} \mathrm{C} $$ turning into water and warming to $$ 25^{\circ} \mathrm{C} $$ ? 42 kj 44 kj 32 kj Rewatch

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**Step 1: Heating the Ice from $$-12^{\circ}\mathrm{C}$$ to $$0^{\circ}\mathrm{C}$$**

The energy required is given by
$$
Q_1 = m \, c_{\text{ice}} \, \Delta T
$$
where  
$$ m = 95 \, \text{g} $$,  
$$ c_{\text{ice}} \approx 2.09 \, \text{J/g}^{\circ}\text{C} $$, and  
$$\Delta T = 0 - (-12) = 12^{\circ}\text{C}$$.

Thus,
$$
Q_1 = 95 \times 2.09 \times 12 \approx 2382.6 \, \text{J}
$$

---

**Step 2: Melting the Ice at $$0^{\circ}\mathrm{C}$$**

The energy required for the phase change (fusion) is
$$
Q_2 = m \, L_f
$$
where  
$$ L_f \approx 334 \, \text{J/g} $$.

Thus,
$$
Q_2 = 95 \times 334 \approx 31730 \, \text{J}
$$

---

**Step 3: Heating the Water from $$0^{\circ}\mathrm{C}$$ to $$25^{\circ}\mathrm{C}$$**

The energy required is
$$
Q_3 = m \, c_{\text{water}} \, \Delta T
$$
where  
$$ c_{\text{water}} \approx 4.18 \, \text{J/g}^{\circ}\text{C} $$ and  
$$\Delta T = 25^{\circ}\text{C}$$.

Thus,
$$
Q_3 = 95 \times 4.18 \times 25 \approx 9927.5 \, \text{J}
$$

---

**Total Energy**

The total energy, $$ Q_{\text{total}} $$, is the sum of the energies from all three steps:
$$
Q_{\text{total}} = Q_1 + Q_2 + Q_3 \approx 2382.6 + 31730 + 9927.5 \approx 44040.1 \, \text{J}
$$
Converting to kilojoules:
$$
Q_{\text{total}} \approx 44.0 \, \text{kJ}
$$

The correct answer is $$44 \, \text{kj}$$.
The total energy required is approximately 44 kJ.

1. How much energy is associated with 95 g of ice at \( -12^{\circ} \mathrm{C} \) turning into water and warming to \( 25^{\circ} \mathrm{C} \) ? 42 kj 44 kj 32 kj Rewatch

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