Question

1.How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 109.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

2. How much energy is released when 42.5 g of water freezes?

3. Suppose that 0.86 g of water condenses on a 75.0 g block of iron that is initially at 24 ∘C.

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.)

Express your answer using two significant figures.

Answer 1

1) Q= mol x H

Q= m x DT x Cp

First heat needed to reach 0ºC:

Q= 13g x 10ºC x 2.09J/gºC= 271.7 J

Then heat needed to change to liquid state:

Q= 13g x 333.55J/g= 4336.15 J

Heat needed to achieve 100ºC:

Q= 13g x 100ºC x 4.184 J/g.ºC= 5439.2 J

Heat needed to convert liquid water to vapour:

Q= 2258.7 J/g x 13g= 29363.1J

Heat needed to reach 109ºC:

Q= 13g x 9ºC x 2.01J/g.ºC= 235.17 J

Total= 271.7 + 4336.15 + 5439.2 + 29363.1 + 235.17= 39645.32 J= 39.6 KJ

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2) Q= 42.5g x 333.55 J/g= 14175.9 J

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3) First let´s find the amount of energy transfered from water to iron:

Q= 0.86g/18g/mol x 44 kJ/mol= 2.1 kJ

the specific heat of iron is 0.449 J/g.ºC, so let´s find the raise of temperature:

T= 2100J/ 75g x 0.449 J/g.ºC= 62.4ºC

Final temperature= 86ºC