VOCAB REVIEW… Potential Energy - energy due to position or composition Kinetic Energy - energy due to motion Temperature - measure of the random motions of the particles in a substance Heat - flow of energy due to a temperature difference Enthalpy - heat associated with a chemical or physical change Law of Conservation of Energy - energy can be converted from one form to another, but never created from nothing or destroyed System – part of the universe that is being analyzed Surroundings – part of the universe that is NOT being analyzed (everything but the system) Specific Heat Capacity - amount of energy required to raise the temperature of 1g of any substance by 1 C calorie – amount of energy required to raise the temperature of 1g H2O by 1 Co joule – smaller unit of energy than the calorie… 1 cal = 4.184 J Endothermic – process that absorbs heat, energy flows into the system from the surroundings Exothermic – process that evolves heat, energy is released into the surroundings from the system Calorimeter - device that uses a change in water temperature to measure the heat flow VOCAB REVIEW…

I’LL INCLUDE ON THE EXAM BECAUSE YOU’RE SOLVING DO NOT FREAK OUT ABOUT ABOUT THIS EXAMPLE CALORIMETRY PROBLEM  IT’S A HARDER ONE THAN I’LL INCLUDE ON THE EXAM BECAUSE YOU’RE SOLVING FOR THE FINAL TEMP, WHICH IS IN BOTH THE q FOR THE METAL AND THE q FOR THE WATER, SO THEY HAVE TO BE SOLVED SIMULTANEOUSLY! FOR AN EXAMPLE OF WHAT IS MORE LIKELY TO BE ASKED ON THE TEST, REVIEW HOW OUR LAB WORKED…I’VE POSTED THE LAB HANDOUT ON THE NEXT SLIDE 

Calorimetry Lab – Specific Heat Capacity The amount of heat (Q) required to raise the temperature of a substance depends on its change in temperature (∆T), its mass (m) and an intrinsic characteristic of the material forming the body called specific heat capacity (s). The heat is calculated from the equation studied in class: q = s x m x ∆T The unit for s is J/gºC and the specific heat of water is 4.184J/gºC. When two bodies in an isolated system, initially at different temperatures, are placed in close contact with each other, they will, in time, come to equilibrium at some common intermediate final temperature. In the end, the colder substance will be warmer than it was initially and the hotter substance will have cooled. Because of energy conservation, the quantity of heat lost by the hot object is equal to that gained by the cold object provided that no heat is lost to the surroundings. This is the theoretical basis for the method of calorimetry through mixture.

A metal sample whose specific heat is to be determined is heated in boiling water to 100oC. It is then quickly transferred to a Styrofoam calorimeter cup which contains a known volume of water at a known temperature. When the metal sample and the calorimeter (including the water) come to equilibrium, the final temperature is measured with a thermometer. The heat lost by the metal sample (qmetal) is equal to the total heat gained by the water (qH2O) plus a small correction factor to account for heat lost to the calorimeter itself, the thermometer and the environment (qcal). This correction factor, qcal will be provided by your instructor because it is different for each type of calorimeter and must be determined experimentally through a process called calibration. Notice that since the metal sample is losing heat, the value of qmetal will be negative and the values of qH2O and qcal will be positive. Viewed more simply in the form of an equation: 0 = qmetal + qH2O + qcal The sum of these three energies is equal to zero because all energy must be accounted for according to the law of conservation of energy. Energy can be transferred but never created or destroyed.

What is the function of the Styrofoam calorimeter? DATA: Consider sig figs! Unknown #1 Unknown #2 Mass of the metal (g)   Volume of water in the calorimeter (mL) Initial temp of water in calorimeter (°C) Initial temperature of metal (°C) Final temp. of mixture in calorimeter °C Pre-Lab Questions: What is the function of the Styrofoam calorimeter? If a 58.0g sample of metal at 100.°C is placed into a calorimeter containing 65.0g of water at 18.0°C, the temperature of the water increases to 21.5°C. Calculate the amount of heat absorbed by the water in Joules. (Note: Assume no heat is lost to the surroundings and no qcal is required)   Determine the identity of the metal by calculating its specific heat (Use table given) How much energy (in kJ) is needed to heat an iron nail with a mass of 7.0g from 25°C until it becomes red hot at 750°C. Show all work.

Calculations: Calculate the temperature change of the water.   Calculate the temperature change of the metal. Calculate the amount of energy absorbed by the water. The specific heat of pure water is 4.184J/g°C, and the density of the water is 1.00g/mL.  Calculate the energy lost by the metal, taking into consideration that it is equal to the energy absorbed by the water plus the correction factor for the calorimeter provided by your instructor. Calculate the specific heat of the metal. Identify the metal. Using the value of the specific heat of the metal, if given by your instructor, calculate your percent error.  Post-Lab Question:   4. Calculate the enthalpy change (∆H) (in Calories) released during the combustion of a peanut that heats up 100.g of water from 20.0°C to 65.0°C. (Note: Food Calories are given in kilocalories where 1 Calorie= 1kcal = 1000cal) Show all work.