5C(21) Effects of temperature on the activity of lipase Aim: To investigate the effects of temperature on the activity of lipase enzyme on milk which contain fats or lipids. Introduction: Enzyme is a kind of biological catalyst that made up of protein. It can speed up the metabolic reactions on various kinds of substances. Like lipase can break down lipid into glycerol or fatty acids in milk. Since enzyme is made up of protein which easily affected by varies temperature. This experiment is
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affects of various temperatures and different light amounts in the different temperatures. We found that as the temperature warmed up‚ the rate of transpiration was higher. When the temperature was at five degrees Celsius‚ the rate of transpiration was very low. When we took out half of the light source and measured the rate of transpiration in the three temperatures we found the same variability of results as earlier where the cold temperature transpired less than the hot temperature but‚ with less
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then press F2 Temperature‚ next press F1 Temperature vs. Time. 3. Half fill a 150-mL beaker with ice and water. Press calibrate to calibrate the temperature probe. When prompted‚ enter 0.0 *C as the actual temperature of the constant temperature bath (ice water). Press Enter. Insert the temperature probe in the ice water. Swirl the temperature probe until the temperature approaches and stabilizes near 0 *C (it may not read exactly 0 *C). Press Enter. 4. Remove the temperature probe from the
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used when the pressure‚ temperature‚ and volume are being measured. The three gas laws are Boyle’s Law‚ Charle’s Law‚ and Gay-Lussac’s Law. The Boyle’s Law is when volume and pressure are being compared. Pressure and volume are inversely proportional‚ because when pressure goes up‚ volume goes down. The Charle’s Law is when volume and temperature are compared. Volume and temperature are also inversely proportional as well. Lastly‚ Gay-Lussac’s Law is when pressure and temperature are compared. Pressure
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|decreases|increases| 4. Two bodies are brought into thermal contact with each other. No thermal energy transfer takes place between the bodies. It may be deduced therefore‚ that the bodies must have the same A. specific heat capacity. B. heat capacity. C. temperature. D. internal energy. 5. An ideal gas expands isothermally‚ doing 2500 J of external work in the process. The thermal energy absorbed by the gas in this process is A. zero. B. less than 2500 J. C. equal to 2500 J. D. more than 2500 J. 6. Which
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A- TYPES OF TEMPERATURE REGULATION 1- The cat‚ human and platypus could be described as true endotherms‚ as they all maintain a relatively constant internal body temperature that is independent of the external temperature. As the environmental temperature rises from 5C to 40C‚ the humans body temperature remains constant‚ the cat’s and the platypus’ increases by around 3▫C. 2- The lizard and echidna are ectotherms as their body temperature fluctuates according to the temperature of the external
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REPRESENTATION OF DATA 1. Pressure versus temperature (P-T) 2. Pressure vs. volume (P-v) 3. Temperature vs. volume (T-v) 4. Temperature vs. entropy (T-s) 5. Enthalpy vs. entropy (h-s) 6. Pressure vs. enthalpy (P-h) The term saturation temperature designates the temperature at which vaporization takes place. For water at 99.6 C the saturation pressure is 0.1 M Pa‚ and for water at 0.1 Mpa‚ the saturation temperature is 99.6 C. If a substance exists
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mixing‚ the temperature of the water is 60ºC. * a. How much did the temperature of the hot water change? How much did the temperature of the cold water change? Compare the magnitudes (positive values) of these changes. 200g of water at 80°C = hot water 100g of water at 20˚C = cold water After mixing the temperature is 60˚C (equilibrium T) Answer: The temperature of hot water changed: 60˚C - 80˚C = -20˚C The temperature of cold water changed: 60˚C - 20˚C = 40˚C The temperature change hot
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Charles Law. Ideal Gas Law Formula : General Gas Equation: PV = nRT Pressure(P) = nRT / V Volume(V) = nRT / P Temperature(T) = PV / nR Moles of Gas(n) = PV / RT where‚ P = pressure‚ V = volume‚ n = moles of gas‚ T = temperature‚ R = 8.314 J K-1 mol-1‚ ideal gas constant. Ideal Gas Law Example: Case 1: Find the volume from the 0.250 moles gas at 200kpa and 300K temperature. P = 200 kPa‚ n = 0.250 mol‚ T = 300K‚ R = 8.314 J K-1 mol-1 Step 1: Substitute the values in the below volume
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cooling water inlet temperature‚ cooling water outlet temperature are vary according to ambient conditions. At the starting of procedure at the starting of experience i.e. on 2:00PM the reading of Air DBT‚ Air WBT‚ cooling inlet‚ cooling water outlet temperature are as follows: At 2:00PM (i.e. starting procedure) all temperature readings are shown in table below. Table 6.1 At 2:00PM (i.e. Starting Procedure) all Temperature Reading Air DBT(0C) Air WBT(0C) Cooling Water Inlet Temperature(0C) Cooling Water
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