Exothermic changes: the system the surroundings. the heat by the surroundings the freedom of motion of the particles which the entropy of the surroundings. therefore qsys 0, qsurr 0, and δssurr 0.

The answers are:

Exothermic changes: the system heats the surrounding. the heat absorbed by

the surroundings increase the freedom of motion of the particles which

increases the entropy of the surrounding. therefore qsys < 0 , qsurr > 0

and delt S > 0 .

the explanation:

the exothermic reaction is a chemical reaction which releases heat, and the

heat is one of its products in the reaction. So the system here released heat

and the surroundings will absorb it.
 
-in the exothermic reaction, the enthalpy in the surroundings will increase,
 
as ΔSsurr = -ΔH/T and when ΔH in the exothermic reaction has a negative

sign( - ) so, ΔSsurr has to be +Ve ( positive sign) ∴ ΔSsurr > 0 .

-and when the system released the heat and the surroundings absorbed it this affects the freedom of motion of the particles and make it increase and this will affect the entropy and make it increase.

- in exothermic reaction when the system released heat and the surroundings absorbed it will increase the thermal disorder in the surrounding which makes 

ΔSsurr >0 and qsys <0 & qsurr >0 as qsurr = - qsys

when ΔSsurr is proportional to the amount of heat transferred

∴ΔSsurr α qsurr  and ΔSurr α - q sys 

so when ΔSsurr has a positive sign in the exothermic reaction so qsurr >0

and qsys<0.

You need to divided twenty four by a dozen them by the temperature it is getting cook by