Energy Day 2 Specific Heat and Calorimetry Youve learned that one calorie, or 4.184 J, is required to raise the temperature of one gram of pure water.

Energy

Day 2 Specific Heat and Calorimetry

• You’ve learned that one calorie, or 4.184 J, is required to raise the temperature of one gram of pure water by one degree Celsius (1°C).

Specific Heat

• That quantity, 4.184 J/(g∙°C), is defined as the specific heat (c) of water.

• The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius.

Specific Heat

• Because different substances have different compositions, each substance has its own specific heat.

• The heat (q) absorbed or released by a substance during a change in temperature depends on

• specific heat of the substance (c) J/gºC

• mass of the substance (m) g

• Amt. of temperature change (∆T) ºC or ºK

Calculating heat evolved and absorbed

You can express these relationships in an equation:

q = c m ∆T

• In the equation

• q = the heat absorbed or released

• c = the specific heat of the substance

• m = the mass of the sample in grams

• ∆T is the change in temperature in °C or °K

Calculating heat evolved and absorbed

• ∆T is the difference between the final temperature and the initial temperature or, Tfinal – Tinitial.

• In the construction of bridges and skyscrapers, gaps must be left between adjoining steel beams to allow for the expansion and contraction of the metal due to heating and cooling.

Calculating Specific Heat

• The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat.

• What is the specific heat of iron?

• You are given the mass of the sample, the initial and final temperatures, and the quantity of heat released.

Calculating Specific Heat

• The specific heat of iron is to be calculated. • The equation that relates these variables can

be rearranged to solve for c. • The first step is to list what you know:

• q = joules of energy released = 114 J • ∆T= 50.4°C – 25.0°C = 25.4°C• m = mass of iron = 10.0g Fe

Calculating Specific Heat

• specific heat of iron, c = ? J/(g∙°C)

• Rearrange the equation q = c m ∆T to isolate c by dividing each side of the equation by m and ∆T.

• Unknown

q = c ∆Tmmm ∆T∆T

Calculating Specific Heat

• Now we can solve the equation using the known values.

q = cm ∆T

q = cm ∆T

114 J

(10.0g) (25.4ºC)= 0.449 J/(g·ºC)

Here is the rearrangedequation we just created

Practice Problem

• A silver bar with a mass of 250.0 g is heated from 22.0°C to 68.5°C. How much heat (q) does the silver bar absorb?From Table 16-2 in your textbook, the specific heat of silver is 0.235 J/(g°C).

Practice cont.• The first step is to list what you know:

– (c) specific heat of Silver is 0.235 J/(g°C ).– (m) mass is 250.0g– (∆T) is the change in temperature

• 68.5°C - 22.0°C = 46.5°C = ∆T

• Then you can plug what you know into the equation and solve!

c m ∆T = q0.235 J/(g°C ) x 250.0g x 46.5°C = q

c m ∆T = q

2730J = q

Here is the equation

Plug in the known values

Calculate the answer!

Measuring Heat

• Heat changes that occur during chemical and physical processes can be measured accurately and precisely using a calorimeter.

• A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

• We often use styrofoam cups as calorimeters in class because they are well insulated .

Determining specific heat

• Suppose you put 125 g of water into a foam-cup calorimeter and find that its initial temperature is 25.6°C.

• You can use a calorimeter to determine thespecific heat of an unknown metal.

Determining specific heat• Then, you heat a 50.0-g sample of the

unknown metal to a temperature of 115.0°C and put the metal sample into the water.

• Heat flows from the hot metal to the cooler water and the temperature of the water rises.

• The flow of heat stops only when the temperature of the metal and the water are equal.

Determining specific heat

• Both water and metal have attained a final temperature of 29.3°C.

• Assuming no heat is lost to the surroundings, the heat gained by the water is equal to the heat lost by the metal.

• This quantity of heat can be calculated using the equation you learned, q = c m ΔT.

• First, calculate the heat gained by the water. For this you need the specific heat of water, 4.184 J/(g∙°C).

Calorimetry• First plug what you know for the water so

you can find qwater

qwater = 4.184J/(g°C) x 125g x (29.3°C -25.6°C)

qwater = 1940J

Since the energy lost by the metal is equal to the energy gained by the water:

qwater = qmetal = 1900J

Determining specific heat

• The heat gained by the water, 1900 J, equals the heat lost by the metal, qmetal, so you can write this equation.

Or more simply put:

Determining specific heat

• Now, solve the equation for the specific heat of the metal, cmetal, by dividing both sides of the equation by m x ∆T.

• Remember that we now need to use the values for m and ∆T that are for the metal sample!

Determining specific heat

• ∆T, is the difference between the final temperature of the water and the initial temperature of the metal

(115.0°C – 29.3°C = 85.7 °C ).

• Substitute the known values of m and ∆T (50.0 g and 85.7 °C) into the equation and solve.

Determining specific heat

• The unknown metal has a specific heat of 0.44 J/(g·°C).

• From the table, you can infer that the metal could be iron.

Using Data from Calorimetry

• A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature increases by 182°C. What is the specific heat of the metal?

• Known • mass of metal = 4.68 g metal • quantity of heat absorbed, q = 256 J • ∆T = 182°C

• Unknown • specific heat, c = ? J/(g·°C)

Using Data from Calorimetry

Energy and Chemical Change: Basic ConceptsEnergy and Chemical Change: Basic Concepts Energy and Chemical Change: Basic ConceptsEnergy and Chemical Change: Basic ConceptsTopic 20Topic 20

• Solve for c by dividing both sides of the equation by m x ∆T.

Using Data from Calorimetry

• Substitute the known values into the equation.

• The calculated specific heat is the same as that of strontium.