Answer :
To determine the energy required to heat 3.50 moles of ice from [tex]\(-4^{\circ} C\)[/tex] to [tex]\(25^{\circ} C\)[/tex] water, we will break down the process into three main steps:
1. Heating the ice from [tex]\(-4^{\circ} C\)[/tex] to [tex]\(0^{\circ} C\)[/tex]:
We start by heating the ice from the initial temperature [tex]\(-4^{\circ} C\)[/tex] up to the melting point of [tex]\(0^{\circ} C\)[/tex]. We use the specific heat capacity of ice for this calculation.
[tex]\[
q_1 = n \times c_{c}(H_2O(s)) \times \Delta T
\][/tex]
Where:
- [tex]\(n = 3.50\)[/tex] moles of ice
- [tex]\(c_{c}(H_2O(s)) = 37.1 \, \text{J/mol°C}\)[/tex] (specific heat capacity of ice)
- [tex]\(\Delta T = 0 - (-4) = 4^{\circ} C\)[/tex]
Calculated energy for heating the ice:
[tex]\[
q_1 = 3.50 \times 37.1 \times 4 = 519.4 \, \text{J}
\][/tex]
2. Melting the ice at [tex]\(0^{\circ} C\)[/tex]:
Next, we convert the ice at [tex]\(0^{\circ} C\)[/tex] to liquid water at the same temperature. This involves using the enthalpy of fusion.
[tex]\[
q_2 = n \times \Delta H_{ts}
\][/tex]
Where:
- [tex]\(\Delta H_{ts} = 6.01 \, \text{kJ/mol}\)[/tex] (enthalpy of fusion, converted to joules so that 1 kJ = 1000 J)
Calculated energy for melting the ice:
[tex]\[
q_2 = 3.50 \times 6.01 \times 1000 = 21035.0 \, \text{J}
\][/tex]
3. Heating the water from [tex]\(0^{\circ} C\)[/tex] to [tex]\(25^{\circ} C\)[/tex]:
Finally, we heat the water from [tex]\(0^{\circ} C\)[/tex] to [tex]\(25^{\circ} C\)[/tex] using the specific heat capacity of water.
[tex]\[
q_3 = n \times c_{p}(H_2O(l)) \times \Delta T
\][/tex]
Where:
- [tex]\(c_{p}(H_2O(l)) = 75.3 \, \text{J/mol°C}\)[/tex] (specific heat capacity of water)
- [tex]\(\Delta T = 25 - 0 = 25^{\circ} C\)[/tex]
Calculated energy for heating the water:
[tex]\[
q_3 = 3.50 \times 75.3 \times 25 = 6588.75 \, \text{J}
\][/tex]
Total Energy Required:
To find the total energy needed for the entire process, we simply add up all the energy values from each step:
[tex]\[
\text{Total energy} = q_1 + q_2 + q_3 = 519.4 + 21035.0 + 6588.75 = 28143.15 \, \text{J}
\][/tex]
Therefore, the total energy required to heat 3.50 moles of [tex]\(-4^{\circ} C\)[/tex] ice to [tex]\(25^{\circ} C\)[/tex] water is 28,143.15 joules.
1. Heating the ice from [tex]\(-4^{\circ} C\)[/tex] to [tex]\(0^{\circ} C\)[/tex]:
We start by heating the ice from the initial temperature [tex]\(-4^{\circ} C\)[/tex] up to the melting point of [tex]\(0^{\circ} C\)[/tex]. We use the specific heat capacity of ice for this calculation.
[tex]\[
q_1 = n \times c_{c}(H_2O(s)) \times \Delta T
\][/tex]
Where:
- [tex]\(n = 3.50\)[/tex] moles of ice
- [tex]\(c_{c}(H_2O(s)) = 37.1 \, \text{J/mol°C}\)[/tex] (specific heat capacity of ice)
- [tex]\(\Delta T = 0 - (-4) = 4^{\circ} C\)[/tex]
Calculated energy for heating the ice:
[tex]\[
q_1 = 3.50 \times 37.1 \times 4 = 519.4 \, \text{J}
\][/tex]
2. Melting the ice at [tex]\(0^{\circ} C\)[/tex]:
Next, we convert the ice at [tex]\(0^{\circ} C\)[/tex] to liquid water at the same temperature. This involves using the enthalpy of fusion.
[tex]\[
q_2 = n \times \Delta H_{ts}
\][/tex]
Where:
- [tex]\(\Delta H_{ts} = 6.01 \, \text{kJ/mol}\)[/tex] (enthalpy of fusion, converted to joules so that 1 kJ = 1000 J)
Calculated energy for melting the ice:
[tex]\[
q_2 = 3.50 \times 6.01 \times 1000 = 21035.0 \, \text{J}
\][/tex]
3. Heating the water from [tex]\(0^{\circ} C\)[/tex] to [tex]\(25^{\circ} C\)[/tex]:
Finally, we heat the water from [tex]\(0^{\circ} C\)[/tex] to [tex]\(25^{\circ} C\)[/tex] using the specific heat capacity of water.
[tex]\[
q_3 = n \times c_{p}(H_2O(l)) \times \Delta T
\][/tex]
Where:
- [tex]\(c_{p}(H_2O(l)) = 75.3 \, \text{J/mol°C}\)[/tex] (specific heat capacity of water)
- [tex]\(\Delta T = 25 - 0 = 25^{\circ} C\)[/tex]
Calculated energy for heating the water:
[tex]\[
q_3 = 3.50 \times 75.3 \times 25 = 6588.75 \, \text{J}
\][/tex]
Total Energy Required:
To find the total energy needed for the entire process, we simply add up all the energy values from each step:
[tex]\[
\text{Total energy} = q_1 + q_2 + q_3 = 519.4 + 21035.0 + 6588.75 = 28143.15 \, \text{J}
\][/tex]
Therefore, the total energy required to heat 3.50 moles of [tex]\(-4^{\circ} C\)[/tex] ice to [tex]\(25^{\circ} C\)[/tex] water is 28,143.15 joules.
