Answer
The total energy required is 480,300 Joules.
Solution
To calculate the total energy required to change 1.00 kg of ice at \(-30.0^{\circ} \mathrm{C}\) to water at \(20.0^{\circ} \mathrm{C}\), we need to consider several steps in the process:
1. **Heating the ice from \(-30.0^{\circ} \mathrm{C}\) to \(0.0^{\circ} \mathrm{C}\)**.
2. **Melting the ice at \(0.0^{\circ} \mathrm{C}\) to water**.
3. **Heating the water from \(0.0^{\circ} \mathrm{C}\) to \(20.0^{\circ} \mathrm{C}\)**.
### Step 1: Heating the Ice
The energy required to heat the ice can be calculated using the formula:
\[
Q_1 = m \cdot c_{ice} \cdot \Delta T
\]
where:
- \(m = 1.00 \, \text{kg}\) (mass of the ice),
- \(c_{ice} = 2.09 \, \text{J/g} \cdot \text{°C} = 2090 \, \text{J/kg} \cdot \text{°C}\) (specific heat capacity of ice),
- \(\Delta T = 0 - (-30) = 30 \, \text{°C}\) (temperature change).
### Step 2: Melting the Ice
The energy required to melt the ice is given by:
\[
Q_2 = m \cdot L_f
\]
where:
- \(L_f = 334,000 \, \text{J/kg}\) (latent heat of fusion for ice).
### Step 3: Heating the Water
The energy required to heat the water can be calculated using the formula:
\[
Q_3 = m \cdot c_{water} \cdot \Delta T
\]
where:
- \(c_{water} = 4.18 \, \text{J/g} \cdot \text{°C} = 4180 \, \text{J/kg} \cdot \text{°C}\) (specific heat capacity of water),
- \(\Delta T = 20 - 0 = 20 \, \text{°C}\) (temperature change).
### Total Energy
The total energy absorbed is the sum of the energies calculated in each step:
\[
Q_{total} = Q_1 + Q_2 + Q_3
\]
Now, let's calculate each of these values step by step.
1. Calculate \(Q_1\):
\[
Q_1 = 1.00 \, \text{kg} \cdot 2090 \, \text{J/kg} \cdot \text{°C} \cdot 30 \, \text{°C}
\]
2. Calculate \(Q_2\):
\[
Q_2 = 1.00 \, \text{kg} \cdot 334,000 \, \text{J/kg}
\]
3. Calculate \(Q_3\):
\[
Q_3 = 1.00 \, \text{kg} \cdot 4180 \, \text{J/kg} \cdot \text{°C} \cdot 20 \, \text{°C}
\]
Now, I will perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
\(1\times 2090\times 30\)
- step1: Rewrite the expression:
\(2090\times 30\)
- step2: Multiply the numbers:
\(62700\)
Calculate or simplify the expression \( 1.00 * 334000 \).
Calculate the value by following steps:
- step0: Calculate:
\(1\times 334000\)
- step1: Calculate:
\(334000\)
Calculate or simplify the expression \( 1.00 * 4180 * 20 \).
Calculate the value by following steps:
- step0: Calculate:
\(1\times 4180\times 20\)
- step1: Rewrite the expression:
\(4180\times 20\)
- step2: Multiply the numbers:
\(83600\)
Now that we have calculated the energy for each step:
1. **Heating the ice from \(-30.0^{\circ} \mathrm{C}\) to \(0.0^{\circ} \mathrm{C}\)**:
\[
Q_1 = 62700 \, \text{J}
\]
2. **Melting the ice at \(0.0^{\circ} \mathrm{C}\)**:
\[
Q_2 = 334000 \, \text{J}
\]
3. **Heating the water from \(0.0^{\circ} \mathrm{C}\) to \(20.0^{\circ} \mathrm{C}\)**:
\[
Q_3 = 83600 \, \text{J}
\]
### Total Energy Absorbed
Now, we can find the total energy absorbed:
\[
Q_{total} = Q_1 + Q_2 + Q_3 = 62700 \, \text{J} + 334000 \, \text{J} + 83600 \, \text{J}
\]
Let's calculate \(Q_{total}\).
Calculate the value by following steps:
- step0: Calculate:
\(62700+334000+83600\)
- step1: Add the numbers:
\(480300\)
The total energy that must be absorbed to change 1.00 kg of ice at \(-30.0^{\circ} \mathrm{C}\) to water at \(20.0^{\circ} \mathrm{C}\) is:
\[
Q_{total} = 480300 \, \text{J}
\]
Thus, the total energy absorbed is **480,300 Joules**.
Answered by UpStudy AI and reviewed by a Professional Tutor
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