Respuesta :
Answer : The amount of heat required is, [tex]1.16\times 10^6J[/tex]
Solution :
The process involved in this problem are :
[tex](1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(182^oC)[/tex]
The expression used will be:
[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = heat required for the reaction
m = mass of ice = 346 g
[tex]c_{p,s}[/tex] = specific heat of solid water or ice = [tex]2.09J/g^oC[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.184J/g^oC[/tex]
[tex]c_{p,g}[/tex] = specific heat of gaseous water = [tex]1.99J/g^oC[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = [tex]333J/g[/tex]
[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization = [tex]2260J/g[/tex]
Now put all the given values in the above expression, we get:
[tex]\Delta H=[346g\times 2.09J/g^oC\times (0-(-80))^oC]+346g\times 333J/g+[346g\times 4.184J/g^oC\times (100-0)^oC]+346g\times 2260J/g+[346g\times 1.99J/g^oC\times (180-100)^oC][/tex]
[tex]\Delta H=1.16\times 10^6J[/tex]
Therefore, the amount of heat required is, [tex]1.16\times 10^6J[/tex]
The total heat required is 1.104 × [tex]10^{6}[/tex] J
In heating mass m = 346g of water from -10°C to 182°C there are the following processes involved:
(i) heating of ice from -10°C to 0°C
ΔQ(1) = msΔT
here we consider s is specific heat capacity of ice = 2.108J/gK
ΔT is change in temperature = 0°C-(-10°C) = 10°C
⇒ ΔQ(1) = 346×2.108×10= 7293.68 J
(ii) conversion of ice to water at 0°C
ΔQ(2) = mL
here L is latent heat of fusion = 330 J/g
⇒ ΔQ(2) = 346×330 = 114180 J
(iii) heating of water from 0°C to 100°C
ΔQ(3) = msΔT
here we consider s as the specific heat capacity of water = 4.184 J/g
⇒ ΔQ(3) = 346×4.184×100 J = 144766 J
(iv) conversion of water to steam at 100°C
ΔQ(4) = mL
here L is latent heat of evaporation = 2260 J/g
⇒ ΔQ(4) = 346×2260 J = 781960 J
(v) heating of steam from 100°C to 182°C
ΔQ(5)= msΔT
here we consider s as specific heat capacity of steam = 1.99 J/g
⇒ ΔQ(5) = 346×1.99×82 J = 56460.28 J
Now the total heat required to heat water from -10°C to 182°C is:
ΔQ(1) + ΔQ(2) + ΔQ(3) + ΔQ(4) + ΔQ(5) = ΔQ
⇒ ΔQ = 1104660 J
ΔQ = 1.104 × [tex]10^{6}[/tex] J is the total heat required to heat water from -10°C to 182°C.
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