A50.0 ml sample of a 1.00 m solution of cuso4 is mixed with 50.0 ml of 2.00 m koh in a calorimeter. cuso4 (1m) + 2koh (2m) cu(oh)2(s) + k2so4 (0.5m). the temperature of both solutions was 20.2 â°c before mixing and 26.3 â°c after mixing. the heat capacity of the calorimeter is 12.1 j/â°c. assume the specific heat and density of the solution after mixing are the same as those of pure water. from this data, calculate theî´h for the process if there is 0.05 mols of cuso4. (energy of the water + energy of the calorimeter)/(1000 x mol)= kj/mol of reaction

- = -

Solution:- Total volume of the solution = 5.0 mL + 50.0 mL = 100.0 mL

Density of solution is same as of water that is 1.0 g per mL. So, the mass of solution, m =

m = 100.0 g

change in temperature, = 26.3 - 20.2 = 6.1 degree C

specific heat, s of water is 4.184 J per g per degree C.

where q is the heat energy. let's plug in the values in the equation:

q = 2552.24 J

energy of calorimeter = change in temperature*heat capacity of calorimeter

energy of calorimeter =  

= 73.81 J

Total heat = 2552.24 J + 73.81 J = 2626.05 J

Let's convert it to kJ and for this we divide by 1000 since, 1000 J = 1 kJ

= 2.626 kJ

0.05 moles of copper sulfate are used so the of the reaction =

=

Since, there is an increase in temperature, the heat is released and so the sign of will be negative.

Hence, = -

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explanation:

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