Using water and the reagents provided in the lab, create two solutions such that when you mix equal amounts together, the resulting solution has an initial temperature of 50'c.

Here’s one of many possibilities.

We MUST know the heat of reaction, ΔH, before we start .

Let’s assume that the reaction is

A + B → Products; ΔH = -80 kJ·mol⁻¹

Let’s assume that you want to get 100 mL of solution that warms from 25 °C to 50 °C.

For the solution

q = mcΔT = 100 g × 4.184 J·K⁻¹mol⁻¹ × 25 K = 10 460 J = 10.46 kJ

The reaction must supply 10.46 kJ.

For the reaction

nΔH = 10.46 kJ

n = 10.46 kJ/80 kJ = 0.131  mol

So, you need 0.131 mol A and 0.131 mol B.

Assume you are using the 3 mol·L⁻¹ solutions.

Then

V = 0.131 mol × (1 L/3 mol) = 0.0436 L = 46.6 mL

You need 46.6 mL of 3 mol·L⁻¹ A + 46.6 mL of 3 mol·L⁻¹ B.

Add 3.4 mL distilled water to 46.6 mL of 3 mol·L⁻¹ A to make 50 mL of A.

Add 3.4 mL distilled water to 46.6 mL of 3 mol·L⁻¹ B to make 50 mL of B.

Mix the two solutions, and you will have 100 mL of a solution at 50 °C .

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante