when a 9.3g sample of solid potassium hyroxide dissolves in 100g of water in a coffee-cup calorimeter the temperature rises from 22.7Cto 43.5C
a) calculate the quantity of heat in (kJ) released in the reaction
b)using the result from part a calculate delta H (in kJ/mol KOH) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

Question

when a 9.3g sample of solid potassium hyroxide dissolves in 100g of water in a coffee-cup calorimeter the temperature rises from 22.7Cto 43.5C
a) calculate the quantity of heat in (kJ) released in the reaction
b)using the result from part a calculate delta H (in kJ/mol KOH) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

jadedry · Answer

A) The specific heat capacity of water is 4.18 J / g ; to find the total energy/heat released, you have to calculate the energy required to increase the temperature of the water + KOH from 22.7 to 43.5C

43.5-22.7 = 20.8C = 20.8K

20.8K * (100g + 9.3g) * 4.18 = 9502.9792 J = 9.502 KJ

B) Calculate the number of moles of KOH you have. Since its molar mass is 56.1 g /mol you have:

9.3/56.1 = 0.1658 Moles . The heat released is proportional to this amount, so to find the Delta H for 1 mole. divide:

9.502 / 0.1658 = 57.31 KJ/ Mol

ac3 · Answer

that's what i got for part A but the back of the book says -8.7kJ

jadedry · Answer

Huh. My bad. 

I can see where they're coming from, just use 100g instead of 109.3g.  (It is just the water after all, I should have guessed.(