Question

Calculate the value of q for each of the experiments listed below. (Assume the heat capacity of water is 4.1801 J/g*C).

Exp1: 2.48g of an unknown salt added to 160. mL of water resulted in a change of temperature of -0.1 C.

Exp2: 2.48g of an unknown salt added to 160. mL of water resulted in a change of temperature of 0.7 C.

Exp3: 2.48g of an unknown salt added to 160. mL of water resulted in a change of temperature of -1.8 C.

Answer 1

salt + H2O ==> solution - heat or
salt + H2O + heat ==> solution. endothermic for #1 and #3. Exothermic for #2. Therefore, for the salt, q is + for 1 and 3 and - for #2.
Do you want q/g? You don't have the molar mass so you can't get q/mol.
qH2O = mass H2O x specific heat H2O x delta T
qH2O = 160 x 4.1801 x 0.1 = about 66.9 J/2.48g = about 27 J/g.

The others are done the same way.

Answer 2

Why did you use 160 instead of 2.48 g ?
and , how do I know #1 and #3 is endothermic and #2 is exothermic ?
thank you

Answer 3

You use the water as a way of knowing how much heat was lost or gained. So the salt changed T from zero c to -1.0 C. Since it cooled the water you know the reaction is endothermic; i.e. it took heat away from the water so the rxn required heat and that is endothermic. Rxns that cause an increase in T means it is giving off heat and that is exothermic.
For #1, the heat taken away from the water is measured by the water.
mass H2O x specific heat H2O x delta T.
The mass H2O = 160g
sp.h. H2O is 4.18 J/g*C
and delta T is 0.1C.
So that gives you q for the 2.48g of the salt that was dropped into the water. q/2.48 = J/gram.