In reaction 3, 4.4 ml of 5.6 m naoh is reacted with an excess hcl. from extrapolation of the temperature curve, the change in temperature is 14°c. calculate the cp of the the calorimete

Answer : The of the the calorimeter is,

Explanation :

First we have to calculate the moles of NaOH.

conversion used for volume : (1 L = 1000 ml)

As we know that, HCl and NaOH are the strong acid and bases. So, the heat of neutralisation is -55.9 kJ/mole.

Now we have to calculate the heat released by neutralization.

As, 1 mole of heat released by neutralization = 55.9 kJ

So, 0.02464 mole of heat released by neutralization =

Now we have to calculate the of the the calorimeter.

where,

Q = heat of reaction = 1.377 kJ

= specific heat capacity of calorimeter =?

= change in temperature =

Now put all the given values in this expression, we get:

Therefore, the of the the calorimeter is,

these nuts

particle   p is an electron and particle q is a neutron