3. What total energy must be absorbed (from the environment) to change 1.00 kg of ice at \( -30.0^{\circ} \mathrm{C} \) on a lake surface in February to water at \( 20.0^{\circ} \mathrm{C} \) in August.

To calculate the total energy absorbed, we first need to consider the processes involved: heating the ice to 0°C, melting the ice, and then heating the resulting water to 20°C. 1. **Heating the ice from -30°C to 0°C**: The specific heat capacity of ice is about 2.09 J/g°C. For 1 kg (1000 g), the energy required is: \[ q_1 = m \cdot c \cdot \Delta T = 1000 \, \text{g} \cdot 2.09 \, \frac{\text{J}}{\text{g°C}} \cdot (0 - (-30)) = 62700 \, \text{J} \] 2. **Melting the ice at 0°C**: The latent heat of fusion is approximately 334 J/g. Thus, the energy required to melt 1 kg of ice is: \[ q_2 = m \cdot L_f = 1000 \, \text{g} \cdot 334 \, \frac{\text{J}}{\text{g}} = 334000 \, \text{J} \] 3. **Heating the water from 0°C to 20°C**: The specific heat capacity of water is about 4.18 J/g°C. Therefore, the energy needed to heat 1 kg of water is: \[ q_3 = m \cdot c \cdot \Delta T = 1000 \, \text{g} \cdot 4.18 \, \frac{\text{J}}{\text{g°C}} \cdot (20 - 0) = 83600 \, \text{J} \] Now, sum the energies: \[ Q_{total} = q_1 + q_2 + q_3 = 62700 \, \text{J} + 334000 \, \text{J} + 83600 \, \text{J} = 480300 \, \text{J} \] Therefore, the total energy that must be absorbed to change 1.00 kg of ice at -30.0°C to water at 20.0°C is **480,300 Joules**.