Calorimetry

Calorimetry

• Calorimetry:• calor (L) + metry (Gr)

• The measurement of heat changes.• food Calories (C) = 1,000 cal

(1 cal = 4.184 J; we use Joules in chemistry)

• 1 cal = heat required to raise the temperature of 1 gram of water by 1oC.

• q sign is the same as before (‘–’ exothermic)

Calorimetry

• Essential Terms:Calorimetry

• Specific Heat (s) amount of heat required to raise the temperature

of 1 gram of substance by 1oC. intensive property

• Heat Capacity (C) amount of heat required to raise the temperature

of a specific system by 1oC. Comparison:

s = heat needed to raise 1 gram of water by 1oC.C = heat needed to raise a glass of ice, tea, &glass by 1oC.

Calorimetry

• Essential Equations: (q=heat; m=mass; D Temp)Calorimetry

• Specific Heat (s) q = m s DT

• Heat Capacity (C) q = C DT(C = m s)

• Heat Exchange (2 substances: A & B)

mA sA DTA

= – mB sB DTB

Calorimetry

Two Types of Calorimeters:Calorimetry

• Constant-Volume Calorimeter• used for combustion reaction

(e.g., Calories in food)

• Constant-Pressure Calorimeter• used for other reactions

(e.g., heat pack: CaCl2 + H2O)

we use this in class

Calorimetry

Example 6.5

A 394-g sample of water is heated from 10.75°C to 83.20°C. Calculate the amount of heat absorbed (in kilojoules) by the water.

Solution

• Specific Heat & Temperature of a Single Substance

CheckSign (+) means heat was absorbed; water was heated; Reaction was endothermic so ‘q’ should be ‘+’

Calorimetry

Example 6.6

A quantity of 1.274 g of naphthalene (C10H8), a pungent-

smelling substance used in moth repellants, was burned in a

constant-volume bomb calorimeter. Consequently, the

temperature of the water rose from 21.49°C to 26.52°C. If the

heat capacity of the bomb plus water was 10.17 kJ/°C,

calculate the heat of combustion of naphthalene on a molar

basis; that is, find the molar heat of combustion.

• Constant-Volume Calorimetry

Calorimetry

Strategy

Knowing mass, DT, and C:

1. How do we calculate the heat absorbed by the calorimeter (qcal) ?

2. What is the heat generated by the combustion of 1.274 g of naphthalene (qrxn) ?

3. What is the conversion factor between grams and moles of naphthalene?

Example 6.6 – Problem Summarym = 1.274 g of C10H8

DT = 21.49°C to 26.52°C; C = 10.17 kJ/°C; q = ? kJ/mol

Calorimetry

1. What is the heat absorbed by the calorimeter (qcal)?

SolutionThe sum of the heat absorbed by the container and the water is equal to the product of the heat capacity and the temperature change. (Assume no heat loss.)

Calorimetry

2. What is the heat generated by the combustion of 1.274 g of naphthalene?

SolutionThe heat absorbed by the calorimeter was released by the chemical reaction.

energy) ofon conservati of (law 0 rxncal qq

Calorimetry

3. What is the conversion factor for naphthalene between grams and moles?

mol 1

g 128

g 274.1

kJ 16.51xqcombustion

kJ/mol 10 x 148.5 3combustionq

Calorimetry

• Measuring Heat Change:

qsys = qcal + qrxn = 0(conservation of energy)

qcal = heat capacity of calorimeter (assume all q is transferred to water)

qrxn = –qcal

mA sA DTA = –mB sB DTB

Calorimetry

Example 6.7A lead (Pb) pellet having a mass of 26.47 g at 89.98°C was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50°C to 23.17°C. What is the specific heat of the lead pellet? Strategy Draw the initial and conditions. (It really does help!)

• Constant-Pressure Calorimetry

Calorimetry

Strategy Treat the calorimeter as an isolated system. We know the masses of water and the lead pellet as well as the initial and final temperatures. Assuming no heat is lost to the surroundings, we can equate the heat lost by the lead pellet to the heat gained by the water. Knowing the specific heat of water, we can then calculate the specific heat of lead.

Summary

• Constant-Pressure Calorimetry

mPb sPb DTPb = – mH2O sH2O DTH2O

Calorimetry

• Constant-Pressure Calorimetry

Solution

Fin

Calorimetry