Thursday, August 20, 2009

Pcos Fibrosis Ovaries

calorimetria

The calorimetry: The part of the thermology who studies the measurement of body heat. All calorimetry calculations are based on the following principles:
when put together two bodies at different temperatures gives the hottest heat to cold until both are at an intermediate temperature of which had at first.
the heat gained by a body is exactly the same as given on the other.
If you have a body mass m1, c1 and specific heat is at a temperature t1 and another mass m2, c2 and specific heat at a temperature that we assume that t2 and t1> t2 by linking them both remain at a temperature intermediate compliments you t1> I> t2. waste heat by the hot body will Qp = m1 c1 (t1-I)
The heat gained by the cold body will Qg = m2 c2 (I-t2) Qp = Qg



In a broad sense calorimetry was developed historically as a technique for manufacturing equipment and procedures to measure the amount of heat released or absorbed in a reaction mechanical, electrical, chemical or otherwise. This discipline, situated within the thermodynamics, has specialized in particular, over time, determining the specific heat of bodies and physical systems.


According to the theories that initiated the study of calorimetry, the heat was a subtle kind of fluid that was produced in the combustion and spent a few bodies to others and can be stored in them more or less cantidad.Posteriormente, we observed that when mechanical work is exerted on a body (to rub or strike, for example), appeared to heat, a fact which contradicted the principle of conservation of energy, and energy disappearing in the form of mechanical work, in addition to being observed the onset of heat without combustion alguna.Benjamin Thompson had been highlighted this fact when he led a boring work canyons, observing that the cooling water holes are heated during processing. To explain, he postulated the theory that heat was a form of energía.Thompson had not shown that conservation of energy in the transformation of work into heat, due to the imprecision in the measures used equipment. Subsequently, Prescott Joule was able to demonstrate experimentally, leading to determine the amount of heat obtained by each work unit consumed, which is 0.239 calories per July work calor.Calorimetría turns entirely on the extent of amount of heat or absorbs a sagging body.



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la calorimetria


HEAT is energy in transit (moving) between 2 bodies or systems, from the existence of a temperature difference between them. Quantity
Heat Units (Q)
units of quantity of heat (Q) are the same units as work (W). Measurement System


Technical System International System (SI) or MKS CGS System


kilogrameter Measurement Unit (Kgm)
Joule (J)
Erg (erg)
Other units used Calorie (cal), kilocalories (Kcal), British Thermal Unit (BTU).
Calorie: the amount of heat needed to raise the temperature of 1 gram of water from 14.5 ° C to 15.5 ° C at a pressure of 1 atmosphere (normal pressure).

Relationship between units

1 kgm = 9.8 J1 J = 107 kgm = 9,8.107 erg erg1
1 cal = 4.186 J 1 kcal = 1000 cal = 10 ³ Btu = 252 cal CAL1

Heat of combustion: the ratio between the amount of heat (Q) provided by certain mass (m) of fuel to be burned, and the mass considered.

heat of combustion Qc ... (In cal / g)
Qc = Q / m

heat capacity of a body: the relationship between the amount of heat (Q) received by a body temperature variation (t) that he experiences.
addition, the heat capacity is a characteristic of each body represents its ability to receive or lose heat by varying its thermal energy.
C. .. heat capacity (in cal / ° C)
C = Q / Q = CX AT AT
specific heat of a body: it is the reason or ratio of heat capacity (C) of a body and mass (m) of the body.
addition, the specific heat should note that is characteristic of substances that constitute the body, while thermal capacity (C) depends on the mass (m) and the material constituting the body.
C. .. specific heat (in cal / g. ° C)
c = C / m C = MXC
We should also note that the specific heat of a substance varies with temperature, increasing when increases, but in our consider that does not change course
The specific heat of water is the exception to this rule, it decreases with increasing temperature in the range of 0 ° C to 35 ° C and grows when the temperature exceeds 35 ° C.
In our course we will consider the specific heat (c) del agua "constante" en el intervalo de 0 °C a 100 °C y es igual a 1 cal / g x °C.
Tabla del calor específico de algunas sustancias

C agua = 1 cal/g.°C
C hielo = 0,5 cal/g.°C
C aire = 0,24 cal/g.°C
C aluminio = 0,217 cal/g.°C
C plomo = 0,03 cal/g.°C
C hierro = 0,114 cal/g.°C
C latón = 0,094 cal/g.°C
C mercurio = 0,033 cal/g.°C
C cobre = 0,092 cal/g.°C
C plata = 0,056 cal/g.°C

Ecuación fundamental de la calorimetría
c= Q/m x AL Q=m x c x AT

Q. .. amount of heat
m. ..
body mass c. .. specific heat of body
Dt ...
temperature variation

Note: For the body temperature increases, must get heat for that tf temperature must be higher than the temperature to, and is called the heat received.
tf> to ® heat received (Q> 0)
To reduce the temperature, it has to lose heat, so that the temperature TF should be lower than the temperature to, and is called the heat transfer. Tf
< to ® calor cedido (Q < 0)
Sensible heat of a body: the amount of heat received or given by a body subject to a variation temperature (Dt) with no change of physical state (solid, liquid or gaseous).
Its mathematical expression is the fundamental equation of calorimetry. Qs = mcΔt

where: At = tf - to
Latent heat of a body: the one that causes the body a change of physical state (solid, liquid or gas) without causing changes in temperature (Dt) is that remains constant. QL = mL

Principles of Calorimetry
1st Principle: When 2 or more bodies with different temperatures are brought into contact, they exchange heat with each other to reach thermal equilibrium.
Then, considering thermally isolated system, "The amount of heat received by each is equal to the amount of heat transferred by the other."
the 2nd Principle: "The amount of heat received by a system during a transformation is equal to the amount of heat transferred by him in the inverse transformation."
Calorimetry - Problems
1 - The heat of combustion of wood is 4 * 10 ³ cal / g. What is the amount of fuel we burn to get 12 * 107 cal?.
2 - The heat of combustion of gasoline is 11 * 10 ³ cal / g. What is the mass of gasoline that we burn to get 40 * 107 cal?.
3 - To heat 800 g of a substance from 0 ° C to 60 ° C were necessary 4.000 cal. Determine el calor específico y la capacidad térmica de la sustancia.
4- Para calentar 2.000 g de una sustancia desde 10 °C hasta 80° °C fueron necesarias 12.000 cal. Determine el calor específico y la capacidad térmica de la sustancia.
5- ¿Cuál es la cantidad de calor necesaria para elevar la temperatura de 200 g de cobre de 10 °C a 80 °C?. Considere el calor específico del cobre igual a 0,093 cal /g °C.
6- Considere un bloque de cobre de masa igual a 500 g a la temperatura de 20 °C. Siendo: c cobre = 0,093 cal /g °C. Determine: a) la cantidad de calor que se debe ceder al bloque para que su temperatura aumente de 20 °C a 60 °C y b) ¿cuál será temperature when they are transferred to 10,000 cal block?
7 - A block of 300 g of iron is 100 ° C. What will your temperature when removed from the 2,000 cal? Knowing that: c iron = 0.11 cal / g ° C.
8 - Sean 400 g of iron at a temperature of 8 ° C. Determine the temperature after giving 1,000 cal. Knowing that: c iron = 0.11 cal / g ° C.
9 - To heat 600 g of a substance at 10 ° C to 50 ° C were needed 2000 cal. Determine the specific heat and thermal capacity of the substance.
10 - What is the amount of heat needed to raise the temperature of 300 g of copper at 20 ° C to 60 ° C?. Where: c = 0.093 cal copper / G ° C.
11 - Sea 200 g of iron at a temperature of 12 ° C. Determine the temperature after giving 500 cal. Where: c iron = 0.11 cal / g ° C.
12 - Transform 20 J in calories.
13 - Transform 40 cal in Joules.
14 - Providing an energy of 10 J to a block of aluminum alloy 5 g, the temperature varies from 20 ° C to 22 ° C. Determine the specific heat of this material.
15 - A thermally insulated container containing 200 g of water, initially at 5 ° C. By means of an agitator, are 1.26 * 104 J delivered to that body of water. The specific heat of water is 1 cal / g ° C, the mechanical equivalent calorie intake is 4.2 J / cal. Consider
negligible heat capacity 16 - 200 g are placed iron at 120 ° C in a container containing 500 g of water at 20 ° C. Since the specific heat of iron equal to 0.114 cal / g ° C and negligible considering the heat absorbed by the recipient. Determine the temperature of thermal equilibrium.
17 - 400 g copper placed at 80 ° C in a container containing 600 g of water at 22 ° C. Determine the thermal equilibrium temperature knowing the specific heat of copper is 0.092 cal / g ° C.
18 - A copper calorimeter contains 62 g 80 g of a liquid at 20 ° C. Calorimeter is placed in an aluminum block of mass 180 ga 40 ° C. Knowing that the temperature thermal equilibrium is 28 ° C, determine the specific heat of liquid. Consider: c Cu = 0.092 cal / g ° C c Al = 0.217 cal / g ° C.
19 - A copper calorimeter contains 25 g 60 g water at 20 ° C. The calorimeter is placed a piece of aluminum mass 120 ga 60 ° C. Since the specific heats of copper and aluminum, respectively equal to 0.092 cal / g ° C and 0.217 cal / g ° C, determine the temperature of thermal equilibrium.
20 - A calorimeter of water equivalent equal to 9 g containing 80 g of water at 20 ° C. A body of mass 50 ga 100 ° C is placed inside the calorimeter. The thermal equilibrium temperature is 30 ° C. Determine specific heat of the body.
21 - is poured inside a calorimeter 150 g of water at 35 ° C. Knowing that the heat initially contained 80 g of water at 20 ° C and the thermal equilibrium temperature is 26 ° C. Determine the water equivalent of the calorimeter.
22 - An iron calorimeter mass equal to 300 g contains 350 g of water at 20 ° C, which is immersed in a lead block of mass 500 g and heated to 98 ° C. The thermal equilibrium temperature is 23 ° C. Since the specific heat of iron equal to 0.116 cal / g ° C. Determine the specific heat of lead.
23 - A copper calorimeter of mass equal to 50 g contains 250 g of water at 100 ° C. An aluminum body temperature of 10 ° C is placed inside the calorimeter. The specific heat of copper is c Cu = 0.094 cal / g ° C and aluminum is c Al = 0.22 cal / g ° C. Knowing that the equilibrium temperature is 50 ° C. What is the body mass of aluminum (approximately)?.
24 - Be a water calorimeter heat capacity 50 cal / g ° C. We took a piece of iron with a mass of 70 g, heated it in a reservoir filled with boiling water vapor, which then introduced into the calorimeter containing 412 g of water at a temperature of 12.4 ° C. Knowing that the final system temperature was 13.9 ° C. Determine the specific heat of iron.
25 - A block of mass 60 g platinum is removed from oven and immediately placed in a copper calorimeter of mass equal to 100 g and containing 340 g of water. Calculate the temperature of the oven, knowing that the initial water temperature was 10 ° C and up to 13 ° C, "when it reached thermal equilibrium?. The specific heat of platinum is 0.035 cal / g ° C and the specific heat of copper is 0.1 cal / g ° C.
26 - A jeweler sold a ring that he said contain 9 grams of gold and 1 g of copper. The ring is heated to 500 ° C (temperature below the melting temperature of gold and copper). Ring is introduced hot water in a calorimeter whose heat capacity is 100 cal / g ° C and the initial temperature is 20 ° C, it was determined that the thermal equilibrium temperature is 22 ° C. The specific heat of gold and copper are 0.09 and 0.031 cal / g ° C, respectively. Determine the masses of gold and copper in the ring.

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calorimetria

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