Respuesta :
The heat is added to 200g ice at 0ºC to make the ice become water. there is no increase in temperature. the heat added is equal to the number of moles of ice melted x molar heat of fusion of ice, 6.01kJ/mole.
200g / 18g/mole = 11.11moles ice
11.11moles x 6.01kJ/mole = 66.78 kJ heat just to melt the ice to water
if you are just looking for the heat to melt the sample of ice, this is where you stop.
if you need to determine the heat to melt the ice AND raise the temperature, this is what you require
the second phase is raising the temperature of the water to 65ºC. this is equal to the mass of the water x specific heat of water x change in temperature
200g x 4.187J/g-ºC x 65ºC = 54431J or 54.431kJ energy to raise the water from 0ºC to 65ºC
the total heat required to melt the ice and then raise the temperature to 65ºC:
66.78kJ + 54.431kJ = 121.21kJ heat
200g / 18g/mole = 11.11moles ice
11.11moles x 6.01kJ/mole = 66.78 kJ heat just to melt the ice to water
if you are just looking for the heat to melt the sample of ice, this is where you stop.
if you need to determine the heat to melt the ice AND raise the temperature, this is what you require
the second phase is raising the temperature of the water to 65ºC. this is equal to the mass of the water x specific heat of water x change in temperature
200g x 4.187J/g-ºC x 65ºC = 54431J or 54.431kJ energy to raise the water from 0ºC to 65ºC
the total heat required to melt the ice and then raise the temperature to 65ºC:
66.78kJ + 54.431kJ = 121.21kJ heat
[tex]\boxed{{\text{121}}{\text{.188 kJ}}}[/tex] of energy is required to melt the sample completely.
Further explanation:
Heat of fusion is the amount of energy needed to melt a solid at its melting point. It is represented by [tex]\Delta {H_{\text{f}}}[/tex].
Step 1: The ice at [tex]{\text{0 }}^\circ {\text{C}}[/tex] is converted to water at [tex]0{\text{ }}^\circ {\text{C}}[/tex].
The formula to calculate the amount of energy is as follows:
[tex]{\text{Q}} = {\text{n}}\Delta {H_{\text{f}}}[/tex] ...... (1)
Here,
Q is the amount of heat required.
n is the number of moles of water.
[tex]\Delta {H_{\text{f}}}[/tex] is the heat of fusion of ice.
The formula to calculate the moles of water is as follows:
[tex]{\text{Moles of water}}=\dfrac{{{\text{Given mass of water}}}}{{{\text{Molar mass of water}}}}[/tex] ...... (2)
Substitute 200 g for the given mass of water and 18 g/mol for the molar mass of water in equation (2).
[tex]\begin{aligned}{\text{Moles of water}}&=\frac{{200{\text{ g}}}}{{{\text{18 g/mol}}}}\\&= 11.11{\text{ mol}}\\\end{aligned}[/tex]
Substitute 11.11 mol for n and 6.01 kJ/mol for [tex]\Delta\text{H}_\text{f}[/tex] in equation (1).
[tex]\begin{aligned}{{\text{Q}}_1}&= \left( {{\text{11}}{\text{.11 mol}}} \right)\left( {\frac{{6.01{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\\&= 66.77{\text{ kJ}}\\\end{aligned}[/tex]
Step 2: The temperature of water is raised from [tex]0{\text{ }}^\circ {\text{C}}[/tex] to [tex]65{\text{ }}^\circ {\text{C}}[/tex] .
The formula to calculate the heat energy of steam is as follows:
[tex]\text{Q}=\text{mc}\Delta\text{T}[/tex] …… (3)
Here,
Q is the amount of heat transferred.
m is the mass of substance.
c is the specific heat of substance.
[tex]\Delta\text{T}[/tex] is the temperature change.
The temperature change for the above change can be calculated as follows:
[tex]\begin{aligned}\Delta\text{T}&=65\;^\circ\text{C}-0\;^\circ\text{C}\\&=65\;^\circ\text{C}\end{aligned}[/tex]
Substitute 200 g for m, [tex]65\;^\circ\text{C}[/tex] for [tex]\Delta\text{T}[/tex] and [tex]4.186\;\text{j/g}^\circ\text{C}[/tex] for c in equation (3).
[tex]\begin{aligned}{{\text{Q}}_2} &= \left( {{\text{200 g}}} \right)\left( {{\text{4}}{\text{.186 J/g }}^\circ {\text{C}}} \right)\left( {{\text{65 }}^\circ {\text{C}}} \right)\\&= 54418{\text{ J}}\\\end{aligned}[/tex]
This energy is to be converted into kJ. The conversion factor for this is,
[tex]1{\text{ J}} = {10^{ - 3}}{\text{ kJ}}[/tex]
Therefore the energy can be calculated as follows:
[tex]\begin{aligned}{{\text{Q}}_{\text{2}}}&= \left( {54418{\text{ J}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ kJ}}}}{{1{\text{ J}}}}} \right)\\&= 54.418{\text{ kJ}}\\\end{aligned}[/tex]
The total amount of energy can be calculated as follows:
[tex]\text{Q}=\text{Q}_1+\text{Q}_2[/tex] ...... (4)
Here, Q is the total energy and [tex]\text{Q}_1[/tex],and [tex]\text{Q}_2[/tex] are the values of energies calculated in first and second step respectively.
Substitute 66.77 kJ for [tex]\text{Q}_1[/tex] and 54.418 kJ for [tex]\text{Q}_2[/tex] in equation (4).
[tex]\begin{aligned}{\text{Q}} &= 66.77{\text{ kJ}} + 54.418{\text{ kJ}}\\&= {\text{121}}{\text{.188 kJ}}\\\end{aligned}[/tex]
Learn more:
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- Find the enthalpy of decomposition of 1 mole of MgO: https://brainly.com/question/2416245
Answer details:
Grade: Senior School
Chapter: Thermodynamics
Subject: Chemistry
Keywords: energy, Q1, Q2, Q, 54.418 kJ, 66.77 kJ, 121.88 kJ, m, c, temperature change, amount of heat, total energy.
