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3 years ago

How much heat is lost when 575 g of molten iron at 1825 K becomes solid at 1181 K and cools to 293 K?

2 answers:

meriva3 years ago

7 0

Answer is: -963,8 kJ.
Q₁ = m(Fe) · C · ΔT₁.
C - specific heat capacity of liquid iron, C(Fe) = 0,82 J/g°<span>C.
</span>m(Fe) = 575 g.
ΔT₁ = 1181 - 1825 = -644°C.
Q₁ = -859306,5 J = -859,3 kJ.
Q₂ = m(Fe) · C · ΔT₂.
ΔT₂ = 293 - 1181 = -888°C.
C - specific heat capacity, C(Fe) = 0,44 J/g°C.
Q₂ = -224664 J = -224,66 kJ.
Q₃ =- heat of fusion, ΔH = 209 J/g.
Q₃ = 120175 J = 120,17 kJ.
Q = Q₁ + Q₂ + Q₃ = -963,8 kJ.

V125BC [204]3 years ago

7 0

Answer:

Total Heat loss = 459.119 J

Explanation:

It is important to realize that there are three processes going on in the problem and each would have it's associated heat.

Amount of heat required to cool down from 1825K to 1181K. = H1

Amount of Heat required for it to change to solid at that temperature. = H2

Amount of Heat for it to cool from 1181K to 293K = H3

Total Heat = H1 + H2 + H3

Calculating H1

H1 = M * C * ΔT

m = 575g

c = specific heat capacity for liquid iron = 0.197 J/gK

ΔT = (1825 -1181) = 644K

H1 = 575 * 0.197 * 644 = 72949.1 J

Calculating H2

The trick here is to realize that the amount of heat needed to melt iron will be equal to the amount of heat given off when molten iron solidifies.

Hence we have;

H2 = mL

where L = Latent heat of fusion = 272 J/g

H2 = 575 * 272 = 156400 j

Calculating H3

H3 = M * C * ΔT

m = 575g

c = specific heat capacity for solid iron = 0.450 J/gK

ΔT = (1181 - 293) = 888K

H3 = 575 * 0.450 * 888 = 229770 J

Total Heat = H1 + H2 + H3

Total Heat = 72949.1 + 156400 + 229770 = 459.119 J

NOTE: the heat capacity of solid and liquid iron are different from each other, as is the case for most substances.

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