In dusteially quicklime is prepared by heating (CaCO3) above 2000°c,

Question

CaCO3(s) = CaO(s) + CO2(g) 177.8kj.
Quicklime melts at 2570°c.exthormic reaction of quick lime with water and small specific heats of both is 0.9j/g.°c.and slake lime is 1.20j/g.°c.
A 500g sample of water is reacted with an equimolar amount of CaO, both at initial of 25°c .what is the final temperature of the product , Ca(OH)2 ? Assum that the product absorbs all the heat released in the reaction? Show calculation step by step.

Answer 1

First, let's calculate the amount of heat released in the reaction of CaO with water:

CaO(s) + H2O(l) → Ca(OH)2(s) ΔH = -65.2 kJ

Since the amount of CaO is equimolar with water, the heat released will be 65.2 kJ.

Next, let's calculate the heat absorbed by the products (Ca(OH)2) to reach the final temperature:

q = m * c * ΔT

where:
q = heat absorbed/released (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)

Now, let's calculate the heat absorbed by the Ca(OH)2:

q = 500g * 1.20 J/g°C * (Tf - 25°C)
q = 600(Tf - 25)

Since the heat released by the reaction is equal to the heat absorbed by the Ca(OH)2:

65.2 kJ = 600(Tf - 25)
65.2 * 1000 = 600(Tf - 25)
65200 = 600Tf - 15000
600Tf = 80200
Tf = 133.67°C

Therefore, the final temperature of the Ca(OH)2 product will be 133.67°C.