For the sample calculations, let’s use the marshmallow as an example. Its initial mass was 0.66 grams and its final mass was 0.36 grams. To calculate the amount burned, subtract 0.36 from 0.66 to get 0.30 grams. (Mass burned = mi- mf).
To find the marshmallow’s change in temperature, use the formula (ΔT = Tf-Ti). Next, subtract the initial temperature, 25 degrees from the final temperature, 29 degrees putting the change in temperature at 4 °C.
To calculate the heat absorbed by the water in calorimeter, use the formula (q = mCΔT). Plug in 50 mL for (m), 4.184 J for (C) and 4 °C for the initial temperature (ΔT), then multiply. The amount of heat absorbed is about 836.8 J and is 199.8 calories and has a value of 0.1998 nutritional calories.
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Since it’s rolled in cheese and fried, which is really unhealthy, the high amount of calories makes sense. We can now conclude that burning the food didn't really make much of a difference in calories, it still shows the cheeto as high in calorie no matter how much was burned off, in this case 1.38 grams was burned. While comparing the nutritional labels of marshmallows and cheetos, the cheetos ended having the highest energy content, as predicted since the fats and carbs are higher than the marshmallows. During the experiment, we only came across a few errors. One our calorimeter did not work so we had to switch to a thermometer for the rest of the lab, our ring stand kept falling over and our lighter stopped working at one point but none of these errors affected our data. Even though nothing went drastically horrible, we didn't have many strengths but had one which was the weighing component because the scale did not malfunction. I believe our results are off on the high end because the data seems to be very accurate. For modifications for a better experiment, I suggest maybe better equipment and more food items. Burning food is quite
This is by using the same mass and realizing that the specific heat of both the regular water and the hot water are the same. In our procedure, 100 mL of hot water was mixed with 100 mL of the regular water; therefore, the masses in Equation 3 cancel out (the densities of the water at different temperatures aren’t exactly the same, but the difference is negligible). This leads to the change in temperature of the hot water equaling the negative change of temperature in the regular water, shown as:
Firstly, when testing temperatures at 30°C and 40°C, the water was. sometimes heated more than needed, so I had to wait until it cooled. down to the required temperature. To avoid this happening, a. thermostatic water bath could have been used, because I could set it. to the required temperature.
When there is a heat exchange between two objects, the object’s temperature will change. The rate at which this change will occur happens according to Newton’s Law of heating and cooling. This law states the rate of temperature change is directly proportional between the two objects. The data in this lab will exhibit that an object will stay in a state of temperature equilibrium, unless the object comes in contact with another object of a different temperature. Newton’s Law of Heat and Cooling can be understood by using this formula:
Investigation of the heat energy produced by combustion of various alcohols. Aim: ---- To investigate how different alcohols produce different amounts of heat energy through combustion. I will be heating water using different alcohols as fuels and measuring the amount of fuel consumed.
water has had equal amounts of time for it to heat up, again I will
-2152.7 x (56.1 / 1.37) = -88150.7 J.mol. 1. H = -88.15 kJ.mol. Hess' law states that: 1"The total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place, provided initial and final conditions are the same.
Abstract: Marshmallows have more Calories per gram. Marshmallows have .2079 J/g℃ and cheese puffs have 1.08x103 J/g℃. My hypothesis was that marshmallows have more Calories per gram and my results confirmed my hypothesis because there is a .2068 J/g℃ difference.
Start with the hot water and first measure the temperature. Record it. 8. Then pour 40 ml into the beaker. You can measure how much water was used by looking at the meniscus.
In calorimetry, in order to measure the amount of heat released or absorbed in reactions, calorimeter is used. The readings are in calories or Btu (British thermal units). There are many types of calorimeters: • Adiabatic Calorimeters • Reaction Calorimeters • Bomb Calorimeters (constant volume calorimeters) • Constant Pressure Calorimeters • Differential Scanning Calorimeters Bomb Calorimeter Bomb calorimeter is a constant volume calorimeter used in bomb calorimetry. It measures the internal energy change of a particular reaction. It is able to withstand large amount of pressure while the reaction is being measured.
In this lab, I determined the amount of heat exchanged in four different chemical reactions only using two different compounds and water. The two compounds used were Magnesium Hydroxide and Citric Acid. Both compounds were in there solid states in powder form. Magnesium Hydroxide was mixed with water and the change in heat was measured using a thermometer. The next reaction combined citric acid and magnesium hydroxide in water. The change in heat was measured as well. For the third reaction citric acid was placed in water to measure the change in heat. In the last reaction, citric acid was combined with water. The heat exchanged was again measured. It is obvious we were studying the calorimetry of each reaction. We used a calorimeter
Temp: Mass of evap. dish: Mass of evap dish+contents: Mass of contents: Solubility g/100cm3 water
(g) Change in mass (g) Methanol 20 170.00 167.08 2.92 Ethanol 20 170.00 167.77 2.23 Propan-1-ol 20 170.00 168.03 1.97 Butan-1-ol 20 170.00 168.16 1.84 Pentan-1-ol 20 170.00 168.24 1.76 Having found these results I then worked out the combustion per mole of alcohol. Alcohol Mass of water heated (g) Heat evolved during reaction (J) Change in mass of burner (g) Combustion of one mole of alcohol (kJ/mole)
- Temperature was measured after and exact time i.e. 1 minute, 2 minutes, 3 minutes.
Sweating and Heat Loss Investigation Aim To find out whether heat is lost faster over a sweaty body compared to a dry body. Apparatus 2 Boiling tubes 47ml max 2 Measuring jug 50ml max A Beaker 250ml max 2 thermometers Paper towels A kettle to boil water A stopwatch 2 magnifying glasses (8x) 2 corks with a small hole through the centre A test tube rack Preliminary work In my preliminary work, I need to find out how much water to use, whether the tissue should be wet with hot/cold water, how often the readings should be taken, how accurate should the readings be, how many readings should be taken and what my starting temperature should be. My results are as follows. Starting temperature of 40°c Time (secs) Wet towel (°c) Dry towel (°c) 30 36 38.9 60 35 38.5 90 34 37.9 120 33.9 37.5 150 33 37 180 32.6 36.9 210 32.3 36.8 240 31 36.5 270 30.4 36 300 30.3 35.9 Starting temperature of 65°c Time (secs) Wet towel (°c) Dry towel (°c) 30 51.1 53 60 48.2 51.9 90 46.4 51 120 46 50 150 44.3 49 180 42.9 48.4 210 42.6 46.9 240 41.7 48 270 40.2 47.5 300 39.3 47 Starting temperature of 60°c Time (secs) Wet towel (°c) Dry towel (°c)
This container must have a value of specific heat capacity so I can calculate heat transferred to it as well. Probably the most conductive container available for use in the classroom is a calorimeter. As well as not wasting energy on the heating of the container, I could also try to stop heat from escaping the top and edges of the container by covering it with a fitting lid. I will try to prevent the wind from blowing the flames in a different direction so all the windows must be shut. HYPOTHESIS More energy is released as more bonds are formed, below is the list of approximate energy required to break and form all bonds involved in burning alcohols.