The problem notes that the change in enthalpy in the solution is +26 kJ/mol. Ignore the heat capacity of NaNO3.
NaNO3(s) -> Na+(aq) + NO3-(aq)
If 50g of NaNO3 is dissolved in 500g of 25.3C water, what will the final temp of the solution be?
2017-11-17 3:37 am
回答 (2)
2017-11-17 3:41 am
C = Q/(m*delta T)
4.18 J/gK = 26000/(550*(T2-25.3))
T2 = 36.6 C
4.18 J/gK = 26000/(550*(T2-25.3))
T2 = 36.6 C
2017-11-17 3:45 am
50 g NaNO3 / 85.0 g/mol = 0.588 mol NaNO3
Heat absorbed = 0.588 mol X 26 kJ/mol = 15.3 kJ absorbed
Since this is absorbing heat, the water must lose -15.3 kJ of heat (or -1.53X10^4 J)
Now, calculate the final temperature of the water
q = m c (T2-T1)
-1.53X10^4 J = 500 g (4.184 J/gC) (T2-25.3)
T2 = 18.0 C
Heat absorbed = 0.588 mol X 26 kJ/mol = 15.3 kJ absorbed
Since this is absorbing heat, the water must lose -15.3 kJ of heat (or -1.53X10^4 J)
Now, calculate the final temperature of the water
q = m c (T2-T1)
-1.53X10^4 J = 500 g (4.184 J/gC) (T2-25.3)
T2 = 18.0 C
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