Question

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If 0.050 mol hydrochloric acid (HCl) is mixed with 0.050 mol of sodium hydroxide (NaOH) in a "coffee cup" calorimeter, calculate the temperature change of 63.7 g of the resulting solution if the specific heat of the solution is 1.00 cal/g·°C and the quantity of heat released by the reaction is 9.3 ×10² cal.

ΔT_w =

I tried to answer it by 14.60 and the answer is wrong so what is ΔT_w =

Answer 1

Dear Student,

In this part of the experiment, the reactions carried out are neutralization reactions between an acid and a base. The heat of reaction or neutralization, q neut, is the negative of the heat gained by the calorimeter

- q neutralization = q cal

qrxn=−qcalorimater=−mCsΔT

qrxn= 9.3 ×10² cal=3.891 KJ

Cs=1.00 cal/g·°C=4.186 KJ/kg∙°C;m=63.7g=0.0637Kg

ΔT= ΔHrxn/-mCs

ΔT= 3.891/(0.0637*4.186)

ΔT= -14.59

The answer is in negative.

Thanks & Best Wishes