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Highly selective Backlog Qs for JEE MainChemistryThermodynamics (C)

A solution of 200 mL of 1 M KOH is added to 200 mL of 1 M HCl and the mixture is well shaken. This rise in temperature T 1 is noted. The experiment is repeated by using 100 mL of each solution and increase in temperature T 2 is again noted. Which of the following is correct?

Options

  1. AT 1 = T 2
  2. BT 2 is twice as large as T 1
  3. CT 1 is twice as large as T 2
  4. DT 1 is four times as large as T 2

Correct answer

A. T 1 = T 2

Step-by-step solution

Heat produced by 200 Meq. of KOH + HCl = 13.7 × 200 1000 k c a l This is used to raise the temperature of 400 mL. Solution say T 1 Thus, 13.7 × 200 1000 = 400 × S × T 1 Now, heat produced by 100 Meq of KOH + HCl = 13.7 × 100 1000 Thus, - 13.7 × 100 1000 = 200 × S × T 2 It is evident that T 1 = T 2 .

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