15) 246 g of hot coffee at 86.0 ° C are placed in a 137 g mug at 20.0 ° C. The s
ID: 1010525 • Letter: 1
Question
15) 246 g of hot coffee at 86.0 ° C are placed in a 137 g mug at 20.0 ° C. The specific heat of coffee is 4.00 J/g ° C, while that of the mug is 0.752 J/g ° C. Assuming no heat is lost to the surroundings, what is the final temperature of the system: mug + coffee?
13) The standard enthalpy of formation for CuSO 4 · 5H 2 O(s) is - 2278.0 kJ/mole at 25 ° C. The chemical equation to which this value applies is:
12) 2 LiOH(s) Li 2 O(s) + H 2 O(l) H ° = 379.1 kJ LiH(s) + H 2 O(l) LiOH(s) + H 2 (g) H ° = - 111.0 kJ 2 H 2 (g) + O 2 (g) 2 H 2 O(l) H ° = - 285.9 kJ Compute H ° in kJ for 2 LiH(s) + O 2 (g) Li 2 O(s) + H 2 O(l)
Explanation / Answer
15) Heat absorbed by mug = specifc heat of mug x temperature rise
= 0.752 J/gC x ( T-20) where T is final temp
Heat lost by cofee = specific heat of cofee x mass of cofee x temp change
= 4 J/gC x 246 g x ( 86-T)
now heat lost by hot cofee = heat gained by mug
hence 4 x 246 x ( 86-T) = 0.752 x ( T-20)
84624 - 984 T = 0.752T - 15.04
T = 85.95 C is final temp of system
13) Cu(s) + 1/8 S8 (s) + (9/2) O2(g) + 5H2(g) ----> CuSO4.5H2O (s) , dH = -2278 KJ/mol
is standard enthalphy fofformation equation. When we deal with standard enthaphy of formation we must form equation from elements in their standard state.
12) 2LiOH --> Li2O + H2O , dH = 379.1 KJ ...........(1)
LiH + H2O ---> LiOH + H2 , dH = -111 KJ .......(2)
2H2 + O2 ---> 2H2O , dH = -285.9 KJ .................(3)
now 2 times eq(2) + eq(1) + eq(3) gives us
2LiH + 2H2O + 2LiOH + 2H2 + O2 ---> 2LiOH + 2H2 + Li2O + H2O + 2H2O
canclelling common terms we get
2LiH + O2 --> Li2O + H2O , dH = 2dH eq2 + dH eq1 + dH eq 3
= 2 ( -111) + 379.1 - 285.9
= -128.8 KJ
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