1.How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 C,
ID: 986141 • Letter: 1
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/gC and that of steam is 2.01 J/gC.
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.
Explanation / Answer
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
------------------------------------------------------------------------------------------------------------------
2) Q= 42.5g x 333.55 J/g= 14175.9 J
---------------------------------------------------------------------------------------------------------------
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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.