5 ml of a dilute HCl solution when mixed with 5 ml of a dilute NaOH solution , initially both present at same temperature , temperature of the mixed solution increased by 1 degree celsius , If 10 ml of both solutions are mixed , maintaining all other conditions same , temperature of the mixed solution will increase by ?

Question

anonymous · Answer

According to what I understood, we use the formula of energy change in neutralization, given by $$E=m*c*DeltaT$$, Where m is the total volume of solution mixed, c is specific heat capacity (of water) and Delta T is the change of temperature. For the first case, we do$$E=(5+5)*4.18*1=41.8J$$
For the second case, we assume that the energy released will be the same as the one of the first case, but now we make DeltaT the subject f the formula, being$$DeltaT=\frac{ E }{ m*c }=\frac{ 41.8 }{ (10+10)*4.18 }=0.5degreesC$$

jfraser · Answer

When you mix 10mL of the solutions, then $twice$ the energy will be released as the 5mL samples because you have twice the number of molecules reacting. The total mass will also be twice as large, so the temperature change will actually remain approximately 1 degree