aluminum from the final temperature of the aluminum. ΔT = Tf - Ti (5 points) ΔT= 99°C – 28°C = 71°C 3. Determine the mass of the water by using the water’s density (specific to the initial temperature) and the volume of the water. Remember‚ density =mass/volume. You can look up the density of the water at your specific temperature at http://www.ncsu.edu/chemistry/resource/H2Odensity_vp.html. (5 points) Density =mass/volume. Density of water according to NCSU: 0.9972995 g/mL Volume of water:
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into the specific heat capacity of a solid Introduction: In this experiment I investigated the specific heat capacity of brass and aluminium‚ in order to do this I arranged a series circuit with a heater‚ the heater would go inside the brass and aluminium and heat it for an amount of time in which I decided‚ I also used the voltmeter in parallel. This arrangement made me able to arrange the equation VIT= mcΔθ‚ since the electrical energy in would be equal to the mass times the specific heat capacity
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1 Heat and Gases Chapter 1 Temperature and Thermometers New Senior Secondary Physics at Work Oxford University Press 2009 1 1 Temperature and Thermometers Practice 1.1 (p. 10) 1 B 2 D 3 A 4 Temperature is a measure of the degree of hotness of an object. 5 (a) On the Celsius temperature scale‚ the lower fixed point is the ice point (0 C) and the upper fixed point is the steam point (100 C). (b) We can reproduce the lower and upper fixed points by using pure melting ice and pure
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Heat Capacity and Coastal Climates: Which Absorbs and Loses Heat Faster––Land or Water? PURPOSE The purpose of this lesson is to investigate the heat capacity of water and land and to explain how the properties of water influence coastal climates. EXPERIMENTAL DESIGN In this experiment‚ you will heat the same volume of water and sand or soil in two different containers. They will compare the temperatures of the water and the sand or soil as each beaker is heated and as each beaker cools to
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Data collection and processing Qualitative observation Before * Blue solution of copper sulfate * Grey zinc powder During * Blue color solution changes to greyish black * Heat is given out‚ so the surroundings become hotter. After * Grey powder remains on the sides of the cup * Precipitate formed at the bottom of the cup List of chemicals and apparatus used Chemicals used Copper sulfate solution‚ 1 mol dm-3 25g of zinc powder Apparatus needed
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g_housesrev3.shtml The specific latent heat of a substance is a measure of how much heat energy is needed to melt or boil it. It is the energy needed to melt or boil 1 kg of the substance. Different substances have different specific latent heats. The specific latent heat of a given substance is different for boiling than it is for melting. The table shows some examples. Latent heats of substances substance specific latent heat of melting kJ/kg specific latent heat of boiling kJ/kg water
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Wrap the wire around a pencil for a slinky effect. Tighten the screws to assure that the wires are fastened. To determine the power P of the immersion heater‚ fill up half of a plastic cup with regular tap water (must be able to cover the wire)‚ and heat it with your immersion heater. By measuring the temperature of water as a function of time‚ and draw a graph‚ you can calculate the power P. The heater is lowered into the water and is connected to an AC power supply (alternating current). At the same
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plants. The uptake of water by plant cells creates a pressure against the rigid ce ll wall. This turgor pressure helps non-woody plants to remain upright. Water has a high specific tension and water molecules have cohesive forces holding them together‚ due to hydrogen bonding between water molecules. This allows water to be pulled through xylem in plants. It is also a transport medium‚ just like in animals‚ substances‚ such
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ΔT. We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. qwater = m × c × ΔT m = mass of water = density x volume = 1 x 26 = 26 grams ΔT = T(mix) - T(water) = 38.9 - 25.3 = 13.6 q(water) = 26 x 13.6 x 4.18 q(water) = 1478 Joules SPECIFIC HEAT: qmetal = -205 J = 15.363 g X c X (27.2 - 100.3 C) c = 0.183 J/gC PART2. Using the formula qmetal = m × c × ΔT‚ calculate the specific heat of the metal. Use the data from
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Questions 1 HKALE 1990 Biology Paper II 1 HKALE 1993 Biology Paper II 1 HKALE 1996 Biology Paper II 1 HKALE 1999 Biology Paper II 1 Suggested Solution 3 HKALE 1990 Biology Paper II 3 HKALE 1993 Biology Paper II 4 HKALE 1996 Biology Paper II 6 Questions HKALE 1990 Biology Paper II 5. (b) Discuss the importance of carbohydrates‚ lipids‚ proteins and roughage in the human diet. (8 marks) Suggested Solution… HKALE 1993 Biology Paper II 6. Based upon its properties‚ discuss
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