Question

What is the final temperature of a solution made by dissolving 0.2 moles of K2SO4 in 250 g water originally at 25 degrees C? Cw= 4.2 J/g*mdegC for water.

Answer 1

Solution :-

The heat of solution of the K2SO4 is 23.8 kJ/ mol

So 0.2 mol K2SO4 * 23.8 kJ / 1 mol = 4.76 kJ

4.76 kJ * 1000 J / 1 kJ = 4760 J

Mass of water = 250 g

Specific heat of water 4.2 J / g C

Initial temperature = 25 C

Lets first calculate the change in the temperature of the water

q=m*c*delta T

4760 J = 250 g * 4.2 J per g C * delta T

4760 J / 250 g * 4.2 J per g C = delta T

4.53 C= delta T

So the change in the temperature of the water is 4.53 C

So the final temperature of the solution is 25 C – 4.53 C = 20.47 C

so final temperature is 20.47 C or we can round it to 20.5 C