If the heat released during condensation goes only to warming the iron block, what is the final temperature (in ∘C) of the iron block? (Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol and a heat capacity for iron of 0.449 J⋅g−1⋅∘C−1.)

Respuesta :

Answer:

[tex]91°C[/tex]

Explanation:

CHECK THE COMPLETE QUESTION BELOW;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree

From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.

Then number of moles =given mass/ molecular mass

Molecular mass of water= 18 g/mol

Given mass= 0.95 g

( 0.95 g/18 g/mol)

= 0.053 moles

Then Heat evolved during condensation = moles of water x Latent heat of vaporization

Q= heat absorbed or released

H=enthalpy of vaporization for water

n= number of moles

Q=nΔH

Q = 0.053 moles x 44.0 kJ/mol

= 2.322 Kj

=2322J

We can now calculate Heat gained by Iron block

Q = mCΔT

m = mass of substance

c = specific heat capacity

=change in temperature

m = 75 g

c = 0.450 J/g/°C

If we substitute into the above formula we have

Q= 75 x 0.450 x ΔT

2322 = 75 x 0.450 x ΔT

ΔT = 68.8°C

Since we know the difference in temperature, we can calculate the final temperature

ΔT = T2 - T1

T1= Initial temperature = 22°C

T2= final temperature

ΔT= change in temperature

T2 = T1+ ΔT

= 68.8 + 22

= 90.8 °C

=91°C

Therefore, final temperature is [tex]91°C[/tex]

The final temperature of the iron block is 91∘C.

Given that;

Heat lost during condensation of the water = Heat gained by iron block

Mass of water(mw) = 0.95 g

Latent heat of vaporization =  Latent heat of condensation(L) = 44.0 kJ/mol

Mass of iron(mi) = 75.0 g

Initial temperature of iron(T1) =  22∘C

Final temperature of iron(T2) = ?

Heat capacity of iron(ci) =  0.449 J⋅g−1⋅∘C−1

So;

mwL = mici(T2 - T1)

Substituting values;

(0.95g/18g/mol) ×  44.0 × 10^3(J/mol) = 75.0(g) × 0.449 J⋅g−1⋅∘C−1 (T2 - 22∘C)

2322.2 = 33.7T2 - 741.4

2322.2 +  741.4 = 37.4T2

T2 = (2322.2 +  741.4)/ 33.7

T2 =91∘C

Missing parts;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

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