PREVNEXTSUBMITMenuResourcesGlossaryNotesClear and return to MenuSearch ResultsFilterSlide TextNotesSearch in:Check to includeSearch...TermsDefinitiondefinitionresourcessearchpauseplayreplaysubmitnextpreviousvolumewelcome to this lesson on calculating heat transfer and change in enthalpy. throughout this lesson we will learn the important variables that are used to calculate changes in enthalpy and how to apply them to different types of chemical processes continue we will begin by looking at a characteristic of substances that explains why some things heat up much faster than others. so grab your swimsuit and sunglasses, you are about to walk out on to the beach on a hot, sunny day next slide specific heat capacity imagine you are at the beach on a hot, sunny day, as you step on to the beach, in your bare feet what happens? how about if you step into the water? click on the beach and then the water to see how our character rosie reacts: next slide specific heat capacity why was rosie happier in the water than on the sand? because the water felt much cooler on rosie's feet than the sand did. why would the water feel so much cooler when they've both been exposed to the sun all day? next the reason is because it takes much more energy (heat) to increase the temperature of the water then it does to increase the temperature of the sand. this difference is due to the specific heat capacity of water vs. the specific heat capacity of sand. next slide specific heat capacity as we just saw, some substances rise in temperature faster than others when exposed to the same amount of heat. the amount of energy that it takes to heat up 1 gram (g) of an object by 1 ºc is known as the specific heat capacity (c) of that object. specific heat capacity is given in units of: j/(g*ºc) and is represented in equations by the symbol: c the greater the specific heat capacity, the longer it will take for an object to raise its temperature. in the case of the sand and the water, the water 4.18 j/(g*ºc) has a much higher specific heat capacity than the sand 0.835 j/(g*ºc) therefore the sand gets hotter much more quickly because it takes much more energy to heat up the water by 1 ºc than the sand calculating heat transfer to determine the amount of heat that is transferred in a chemical process. q = m x c x δt where: q = amount of heat absorbed or released by the system. units = (j) or (kj) or (j/mol), (kj/mol). m = mass of the substance you are measuring. units = (g). c = specific heat capacity of the substance you are measuring. units = j/(g*ºc). δt = the temperature change of the system (final temperature - initial temperature). q is positive when heat is gained by the system and negative when heat is lost by the system. since q is a measure of the change in heat, it is equivalent to δh. continue lets try a sample problem: how much heat is released when 20.0 g of copper, which has a specific heat capacity of 0.385 j/(g*ºc) is cooled from 85.0 ºc to 25.0 ºc? try to solve this on your own before moving on to the next slide for the solution. next slide calculating heat transfer how much heat is released when a 20.0 g piece of copper, which has a specific heat capacity of 0.385 j/(g*ºc), is placed in a sealed cup of water and cooled from 85.0 ºc to 25.0 ºc? continue step 1: identify the variables. q = m x c x δt q = ? m = 20.0 g c = 0.385 j/(g*ºc) δt = (tfinal - tinitial) = (25.0 ºc - 85.0 ºc) = -60 ºc step 2 85 ºc 25 ºc step 2: solve for the unknown q = m x c x δt q = 20.0 g x 0.385 j/(g*ºc) x (-60.0 ºc) q = -462 j step 3 step 3: solution q = - 462 j, since the value of q is negative, this means that the system released heat. therefore, 462 j of heat is released when 20.0 g of copper is cooled from 85.0 ºc to 25.0 ºc. next slide 85 ºc 25 ºc 462 j 85 ºc 25 ºc calculating heat transfer we just saw in the last example that when a heated piece of copper was placed in a sealed container of water, the value of q for the copper was - 462 j. if we assume that all of this heat was transferred to the water, we can also say that + 462 j of heat was absorbed by the water. in other words: qsystem = - qsurroundings - 462 j + 462 j continue we can use the equation, q = m x c x δt, and the relationship, qsystem = - qsurroundings to determine the δh of a system by measuring the changes in temperature of the surroundings, provided we know the mass and specific heat capacity of the surroundings. next slide measuring δh : calorimetry we just learned how to calculate heat transfer using the equation q = m x c x δt, the next step is learning how to apply that equation to measure δh values for different types of chemical reactions. to measure this, scientists have developed a technique known as calorimetry which uses a device known as a calorimeter to measure the precise values of heat absorbed or released in a chemical reaction or change of state. continue in a calorimeter, the heat absorbed or released by a system is measured by a change in temperature in the surroundings. the image to the right represents a simple, constant pressure calorimeter called a coffee cup calorimeter. in this type of calorimeter, a chemical reaction (system) takes place in water (surroundings). the δh is calculated by using the temperature change as well as the mass and specific heat capacity of water. styrofoam cups styrofoam lid stirrer thermometer water next slide calorimetry sample problems lets try an example to calculate the δh for the dissolution of kbr in water. when a substance dissolves in water, there is a change in enthalpy that can be calculated by a calorimeter. the procedure for this involves adding a known mass of kbr to a known mass of water with a known initial temperature and simply observing the temperature change. step 1 step 1: record initial temperature of water. step 2 step 2: add the sample. step 3 step 3. observe the temperature change, while stirring the solution constantly. when the temperature stops changing, record the final temperature. step 4 step 4. plug in the values for initial temperature, final temperature, specific heat capacity of water and mass of water used in the experiment into the equation: q = m x c x δt. this will give you the amount of heat transferred to the water. reverse the sign to give you the amount of heat transferred from the system. this value of q for the system is its δh. next slide calorimetry problems a student adds 125 g of water to a coffee cup calorimeter and records its initial temperature at 23.4 ºc. she then adds 10.5 grams of solid potassium bromide (kbr) to the water in the calorimeter and begins stirring. she notices the temperature begins to drop. when the kbr is fully dissolved, she records the final temperature at 20.3 ºc. calculate the heat transfer for the system (kbr dissolving) as well as the δh of the system and determine if the reaction is endothermic or exothermic. step 1 step 1: identify the variables. qsurroundings = m x c x δt q = ? mwater = 125.5 g cwater =4.18 j/(g*ºc) δt = (20.3 ºc - 23.4 ºc) = -3.1 ºc step 2 step 2: solve for the unknown qsurroundings = m x c x δt qsurroundings = 125 g x 4.18 j/(g*ºc) x (-3.1 ºc) qsurroundings = -1620 j note: round to the nearest significant digit step 3 step 3: solution qsurroundings = - 1620 j. qsystem = - qsurrounding; therefore qsystem = + 1620 j. in this problem, the system absorbed 1620 j of energy. therefore δh of the system is +1620 j and this reaction is endothermic. so we can say that when 10.5 g of kbr is dissolved in 125 g of water, the δhsystem is +1620 j or 1.62 kj. next slide molar enthalpy in the previous problem we determined the δhsystem was +1620 when 10.5 g of kbr was dissolved in water. the δh when a substance dissolves is called the: enthalpy of solution. often times, we want to know what the molar enthalpy of solution of a given system is (δhsol). the molar enthalpy is the δh that occurs when 1 mole of a substance reacts. to do this we need to determine the number of moles in 10.5 grams then calculate what the δh would be for 1 mole of kbr. step 1 step 1: determine the molar mass (m) of kbr. we use the periodic table to calculate this. mk = 39.1 g/mol; mbr = 79.9 g/mol mkbr = (39.1 + 79.9) = 119.0 g/mol step 2 step 2: calculate the number of moles (n) recall the formula for calculating number of moles: n = mass of sample / molar mass n = 10.5 g / 119.0 g/mol n = 0.0882 mol step 3 step 3: calculate molar enthalpy of solution (δhsol) δhsol = δh/n δhsol = +1620 j / 0.0882 mol δhsol = 18367 j / mol step 4 therefore, the molar change in enthalpy for the dissolution reaction of kbr is 18367 j/mol next slide ok, so we learned quite a bit in this lesson, lets recap quickly. to measure the amount of heat that is transferred in a chemical process, we can use the equation: q = m x c x δt. next scientists use a device called a calorimeter to measure heat transfer by calculating the value of q for water in which a chemical process is taking place. because we know the specific heat capacity of water we are able to easily calculate heat transfer in water by measuring its change in temperature due to the chemical process next the heat that is transferred to or from the water in a calorimeter the opposite of the amount of heat that is transferred to or from the system. therefore if we calculate q for the water (surroundings) we know the value of qsystem will be that same magnitude but opposite sign. in other words qsystem = - q surroundings finish <html><textformat leading='1' tabstops='[48, 96]' leftmargin='0' indent='0'>Classroom-themed layout features a retro blackboard placeholder. Use this fun and informal layout with your favorite handwriting fonts to complete the schoolhouse-inspired design.</textformat></html><html><textformat leading='1' tabstops='[48, 96]' leftmargin='0' indent='0'>Classroom-themed layout features a retro blackboard placeholder. Use this fun and informal layout with your favorite handwriting fonts to complete the schoolhouse-inspired design.</textformat></html>