Lab 6-Food Calorimetry Lab Page 50 Reactions that combine a fuel source of some
ID: 1038599 • Letter: L
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Lab 6-Food Calorimetry Lab Page 50 Reactions that combine a fuel source of some type with oxygen are called combustion reactions. The products of complete combustion are CO occurs it gives off energy in the form of heat energy and usually light energy as well. When a gas and H20 vapor. When a combustion action liberates heat it is said to be exothermic. These reactions feel hot to the observer. Other reactions take in heat and are said to be endothermic. These reactions feel cold to the observer. The heat given off or taken in is measured using various units such as Joules (W), calories (cal), or Food Calories (Cal). If we look at the label on a package of food, we will see information about the calories in a serving of the food item. Today we are going to use a can and water as a calorimeter to measure the energy in a snack food. Calorimetry is based on the concept of Conservation of Energy. The First Law of Thermodynamics states that the heat energy lost by one object is gained by another object. Heat (a) can be transferred when there is a the object at a higher temperature are moving faster than particles in an object at a lower temperature. If the two objects contact on e another, the faster particles smash into the slower les and transfer some energy with each collision, until the two objects reach equilibrium. The colder object will be warmer than it was at first, and the object that was hotter will be cooler than it was at first. Measuring transfers of heat energy Every substance holds a specific amount of heat per gram (or per mole). The specific heat (Cop) of a substance is the amount of heat energy needed to raise one gram of the substance by one degree Celsius. Since water is so common, its specific heat was used as the definition of the calorie. 1 calorie is the amount of heat eneray needed to raise one grom of water 1°C This means that the specific heat of water is 1 cal/ec (or 4.18 J/gK). These two specific heats of water are equivalent just using different units. In this lab, we will burn a food item and use it to heat water in a can above it. We can use the change in temperature of the water to calculate how much heat the burning food released. When we are talking about food, however, we are actually using kilocalories, aka Calories (with a capital c) There are 1000 calories in a Calorie Sp Heat -wAT M- mass (8) Q- Heat (U) The equation to calculate specific heat is: Where Sp Heat of water 4.18 J/gK AT = TFinal-Tinitial (K) Do not ingest any of the snack foods used in this lab! Once a food substance is brought into the lab it is considered contaminated. Safety Precautions!!!! 1. Food Calorimetry LabExplanation / Answer
For determination 1
= 94.5 – 69.8 = 24.7 oC
For determination 2
= 122.7 – 90.3 = 32.4 oC
2. Heat gained by the water
Sp = Q/M?t
Q = M ? Sp ? ?T
where Q = heat energy absorbed by water
M = mass of water being heated
Sp = specific heat of water = calories required to raise temperature of 1 gram of water 1 C o = g ? C o 1 calorie
?T = change in temperature of water = final temperature ? initial temperature
For determination 1
Q = 115.72g * 4.18 J/g.K *24.7K
= 11947.6 J
For determination 2
Q = 115.72g * 4.18 J/g.K *32.4 K
= 15672.2 J
3.Convert the Joules to food Calories
For determination 1
Q = 11947.6/4.18 = 2858.3 calories = 2.86 Calories
For determination 2
Q = 15672.2/4.18 = 3749.3 calories = 3.79 Calories
4. Frito burned
Mass of unburned – mass of burned = mass burned up
For determination 1
= 8.369-7.06 =1.309
For determination 2
= 8.58 – 6.91 = 1.67
5.Frito’s energy content in Calories per gram
Calories / g of Frito burned
For determination 1
= 2.86 C / 1.309g = 2.18
For determination 2
= 3.79 C / 1.67g = 2.25
6. Mean from two determinations
= (2.18+2.25)/2 = 4.43/2 = 2.215
7. Accepted value should be taken from the package (from nutritional info) of Frito.
8. Percentage collected = (2.215 / accepted value(Ans #7) ) *100
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