Problems in calorimetry

Example 1 on Calorimetry

There is an increase in 1⁰ C for every 1560 J for a certain material. A 0.1 gram of quinone (molar mass = 108.1 g/mole) when burnt, resulted in an increase in temperature from 22⁰C to 25.2⁰C. Calculate the molar heat of combustion for quinone.

Solution:

In this case, the change in heat is given by Q = m*c* delta T.

Hence the value of “Q” for the material is given as Q = 1560 (25.2 – 22) J/⁰C

Where delta T = change in temperature

= final temperature – initial temperature

= (25.2 – 22) = 3.2 ⁰C

Hence Q = 1560 * 3.2 = - 4992 J/0C = -5KJ/0C

Number of moles = mass / molecular mass = 0.1 / 108.1 = 9.25 * 10^{- 4} moles

Hence molar heat = - 5 / (9.25 * 10 ^-4)

= 5.4 * 10 ^-3 KJ/mole of quinone. (Answer)

Sub Topics

Example 2 on Calorimetry

Example 2 on Calorimetry

0.5 moles of a certain acid HX was exactly neutralized by KOH. The temperature rose from 20⁰C to 24.7⁰C. Find the exact concentration of the acid HX.

(Given that- the molar heat of neutralization = -80KJ/mole)

Solution:

Heat of neutralization of 0.5 moles of KOH = 0.5 * 80 = -40KJ/mol.

Hence Q = -40KJ = -4000J

We know that Q = m * c * delta T

Or m = Q /(c * delta T) = 40 / (4.19)*(24.7-20)

Or m = 40 / 19.693 = 2.031 kg = 2031 g of water.

The acid concentration and the number of moles are equal, hence the quantity of HX and NaOH ( the volumes) will be equal.

2031 g = 2031 ml, and that is made up from 2031 /2 = 1015.5 ml of acid

= 1.0155 l of acid.

Hence the concentration of the acid will be given as

Number of moles / volume (in liters)

= 0.5 / 1.0155 = 0.4923 M (Answer)