In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 5.10 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol .

Answer : The final temperature of the solution in the calorimeter is,

Explanation :

First we have to calculate the heat produced.

where,

= enthalpy change = 82.8 kJ/mol

q = heat released = ?

m = mass of = 5.10 g

Molar mass of = 110.98 g/mol

Now put all the given values in the above formula, we get:

Now we have to calculate the final temperature of solution in the calorimeter.

where,

q = heat produced = 3.80 kJ = 3800 J

m = mass of solution = 100 + 5.10 = 105.10 g

c = specific heat capacity of water =

= initial temperature =

= final temperature = ?

Now put all the given values in the above formula, we get:

Thus, the final temperature of the solution in the calorimeter is,

b) mutualism

hope this : )

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2