Respuesta :
Answer:
∑Q(TTL) = 347 Kj. = 82.8 Kcal. (3 sig. figs.)
Explanation:
Draw a typical heating curve for water (see figure below) and label each section with data needed to calculate the amount of heat flow in the specified section.
Super-imposing given data onto a trace of a typical heat flow chart for water finds there are three segments of the heating curve that can utilize the given data to calculate heat flow for each segment. With all three heat flow quantities, the total heat quantity associated with this problem is marked in orange. Calculate the heat quantity associated with the designated segments and add to obtain total heat flow for the transitions listed.
Point E to D => Q₄ = m∙ΔHᵥ = (115g)(540 cal/g) = 62,100 cals.
Point D to C => Q₃ = m∙c∙ΔT = (115g)(1 cal/g∙˚C)(100˚C) = 11,500 cals.
Point C to B => Q₂ = m∙ΔHₓ = (115g)(80cal/g) = 9,200 cals.
Total Heat Flow (∑Qₙ) = Q₄ + Q₃ + Q₂ = 62,100 cals. + 11,500 cals. + 9,200 cals. = 82,800 cals.
= (82,800 cals.)(4.184 joules/cal.) = 346,435 joules = 346.435 Kj ~ 347 Kj (3 sig. figs.) _____________________________________________________
m = mass = 115g
c = specific heat of liquid water = 1 calorie/gram·°C = 4.184 j/g·°C
ΔT = temperature change in degrees Celsius
ΔHᵥ = heat of vaporization = 540 cals./gram
ΔHₓ = heat of crystallization = 80 cal./gram
Qₙ = heat flow quantity per specific segment (calories or joules)
∑Qₙ = total heat flow
Heating curve for water:
Note in the diagram that only two formulas are used.
Q = m·c·ΔT => heating or cooling the pure condensed state. The segments demonstrate temperature change.
Q = m·ΔHₙ => heat flow during phase change. Note, in these segments of the heating curve two phases are in contact. That is solid/liquid or liquid/gas phase substances. Also note, in these segments, when two phases are in contact no temperature change occurs. Examples, ice water remains at a constant temperature until all ice is melted or all of the liquid water is frozen depending upon the direction of heat flow. The same is true for boiling water in that when two phases are in contact (liquid/gas), temperature remains constant. The portions of the heating curve designating phase transitions are horizontal and are defined by the equation Q = m·ΔH, while the curve segments that are only one phase demonstrate temperature change and are defined by the equation containing temperature change, Q = m·c·ΔT.
The total energy released during this whole process was 346.4 kiloJoules.
Explanation:
Given:
115 g of steam at 100°C condenses, cools, and falls as snow at 0.0°C.
To find:
The energy released during this whole process.
Solution:
In the given three stages will appear:
- [tex]H_2O(g)\rightarrow H_2O(l) (at 100^oC)[/tex]
- [tex]H_2O(l)(at 100^oC)\rightarrow H_2O(l) (at 0.0^oC)[/tex]
- [tex]H_2O(l)(at 0.0^oC)\rightarrow H_2O(s) (at 0.0^oC)[/tex]
1) [tex]H_2O(g)\rightarrow H_2O(l) (at 100^oC)[/tex]
Phase change will occur at 100°C, that is from gas to liquid.
Energy released during the phase change of water from gas to liquid =[tex]Q_1[/tex]
Latent heat of vaporization of water =[tex]\Delta H_{vap}=2.260 kJ/g[/tex]
Latent heat of condensation of water =[tex]\Delta H_{con}=- \Delta H_{vap}=2.260 kJ/g=-2.260 kJ/g[/tex]
Mass of steam = 115 g
[tex]Q_1=115 g\times \Delta H_{con}\\=115g\times -2.260 kJ/g=-259.9 kJ[/tex]
2) [tex]H_2O(l)(at 100^oC)\rightarrow H_2O(l) (at 0.0^oC)[/tex]
Mass of water = m = 115 g ( steam converted into water)
Initial temperature of water = [tex]T_i=100^oC[/tex]
Final temperature of water =[tex]T_f=0.0^oC[/tex]
The specific heat of water = c = 8.186J/g°C
Energy released during this stage = [tex]Q_2[/tex]
[tex]Q_2=m\times c\times (T_f-T_i)\\=115g\times 4.186 J/g^oC\times (0.0^oC-100^oC)\\=-48,139 J= -48.139 kJ\\(1 J=0.001kJ)[/tex]
3)[tex]H_2O(l)(at 0.0^oC)\rightarrow H_2O(s) (at 0.0^oC)[/tex]
Water freezes at 0°C ,phase change will occur at 0.0°C, that is from liquid to solid.
Energy released during the phase change of water from liquid to solid=[tex]Q_3[/tex]
Latent heat of fusion of water =[tex]\Delta H_{fus}=-334J/g[/tex]
Mass of water = 115 g
Energy released during this stage:
[tex]Q_3=115g\times \Delta H_{fus}\\115\times -334 J/g=-38,410 J\\=-38,410 J=-38.410 kJ\\(1 J=0.001kJ)[/tex]
Total energy released during this whole process = Q
[tex]Q = Q_1+Q_2+Q_3\\Q=(-259.9 kJ)+( -48.139 kJ)+(-38.410 kJ)\\=-346.449 kJ\approx -346.4 kJ[/tex]
(negative sign indicates that heat energy is released))
The total energy released during this whole process was 346.4 kiloJoules.
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