Presentation on theme: "Day 2 Specific Heat and Calorimetry"— Presentation transcript:
1Day 2 Specific Heat and Calorimetry EnergyDay 2 Specific Heat and Calorimetry
2Specific HeatYou’ve learned that one calorie, or J, is required to raise the temperature of one gram of pure water by one degree Celsius (1°C).That quantity, J/(g∙°C), is defined as the specific heat (c) of water.
3Specific HeatThe 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.Because different substances have different compositions, each substance has its own specific heat.
4Calculating heat evolved and absorbed The heat (q) absorbed or released by a substance during a change in temperature depends onspecific heat of the substance (c) J/gºCmass of the substance (m) gAmt. of temperature change (∆T) ºC or ºKYou can express these relationships in an equation:q = c m ∆T
5Calculating heat evolved and absorbed In the equationq = the heat absorbed or releasedc = the specific heat of the substancem = the mass of the sample in grams∆T is the change in temperature in °C or °K∆T is the difference between the final temperature and the initial temperature or, Tfinal – Tinitial.
6Calculating Specific Heat 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.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?
7Calculating Specific Heat You are given the mass of the sample, the initial and final temperatures, and the quantity of heat released.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°Cm = mass of iron = 10.0g Fe
8Calculating Specific Heat Unknownspecific 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.q=cm∆Tm∆Tm∆T
9Calculating Specific Heat Here is the rearrangedequation we just createdq=cm∆TNow we can solve the equation using the known values.114 Jq=c= J/(g·ºC)(10.0g)m(25.4ºC)∆T
10Practice ProblemA silver bar with a mass of 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 J/(g°C).
11Practice cont. The first step is to list what you know: (c) specific heat of Silver is J/(g°C ).(m) mass is 250.0g(∆T) is the change in temperature68.5°C °C = 46.5°C = ∆TThen you can plug what you know into the equation and solve!
12c m ∆T = q 0.235 J/(g°C ) x 250.0g x 46.5°C = q 2730J = q c m ∆T = q Here is the equationc m ∆T = qc m ∆T = q0.235 J/(g°C ) x 250.0g x 46.5°C = qPlug in the known values2730J = qCalculate the answer!
13Measuring HeatHeat 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 .
14Determining specific heat You can use a calorimeter to determine thespecific heat of an unknown metal.Suppose you put 125 g of water into a foam-cup calorimeter and find that its initial temperature is 25.6°C.
15Determining 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.
16Determining 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, J/(g∙°C).
17CalorimetryFirst plug what you know for the water so you can find qwaterqwater = 4.184J/(g°C) x 125g x (29.3°C -25.6°C)qwater = 1940JSince the energy lost by the metal is equal to the energy gained by the water:qwater = qmetal = 1900J
18Determining 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:
19Determining 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!
20Determining 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 = °C ).Substitute the known values of m and ∆T (50.0 g and 85.7 °C) into the equation and solve.
21Determining 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.
22Using 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?Knownmass of metal = 4.68 g metalquantity of heat absorbed, q = 256 J∆T = 182°CUnknownspecific heat, c = ? J/(g·°C)
23Using Data from Calorimetry Energy and Chemical Change: Basic ConceptsTopic20Using Data from CalorimetrySolve for c by dividing both sides of the equation by m x ∆T.
24Using Data from Calorimetry Substitute the known values into the equation.The calculated specific heat is the same as that of strontium.