How To Find Heat Gained By Water

SPECIFIC HEAT AND Rut OF FUSION

Function I. Specific estrus

HEAT EXCHANGE: When ii or more objects at different temperatures are brought together in an isolated environment, they eventually reach the same temperature by the process of heat exchange. That is, warmer materials transfer heat to colder materials until their temperatures are the same. The energy which is being transferred is referred to equally internal energy, energy associated with random molecular move on the microscopic calibration. This energy can be divided into kinetic energy and the potential energy arising from the intermolecular attractive forces in the material. If we accept 2 isolated substances and assume no loss to the environment, then the conservation of energy principle implies that the energy lost by ane substance must be gained by the other. Letting Q stand for the quantity of heat, this idea may exist expressed as

(1)

Temperature is a measure of the boilerplate kinetic energy of the random molecular motion. Equally a body gains or loses kinetic energy its temperature will increase or subtract. The temperature change is symbolized by DT where

(2)

Although the internal energy possessed by an object is directly proportional to its mass, it does not follow that two objects of the same mass and temperature have the same amount of internal energy. Temperature reflects just the kinetic energy portion of the internal free energy, so a substance with a greater fraction of its internal energy in the form of potential energy will have a greater internal energy at a given temperature. Thus a gram of water at 50°C volition accept a much larger internal free energy than a gram of copper at l°C and it will accept a much larger free energy input to heat the h2o to 60°C than to produce the same temperature change in copper. This belongings is reflected in the quantity called the specific oestrus (S). The specific oestrus is divers every bit the quantity of heat required to raise the temperature of 1 gram of a substance ane°C.

Taking all of the above into consideration, the quantity of heat lost or gained past a body tin can be calculated by the relation

(3)

where M is the mass of the substance.

* Internal energy can exist measured in calories. The dietary Calorie is a kilocalorie or 1000 calories.

LABORATORY Exercise: The specific estrus of a metal sample will be determined. A summary of the experiment follows, and the appliance video accessible from the right sidebar volition show you the pieces of equipment involved.

Heated metal is poured into an insulated container of cool h2o. Since care will be taken to foreclose estrus from entering or escaping the container, information technology is causeless that the amount of free energy lost by the hot metal is gained past the water.

From relationships (ii) and (iii) the corporeality of heat gained by the h2o can be computed. Since this must be the amount of oestrus lost by the metal, relationship (3) can be practical to the metal sample. When substitution of the known quantities is done, the remaining unknown quantity is the specific estrus of the metal, which is then calculated from the equation.

Process:

1. Fill up the boiler about 2/3 total of water. Begin heating immediately.

2. Place a known mass of metal in the boiler cup as follows:

a. Determine the mass of the empty boiler cup.

b. Fill the boiler loving cup near ii/3 full of the metal material provided and redetermine the mass.

3. Carefully identify the boiler cup into the boiler so the metal tin begin heating.

4. Using the graduated cylinder, pour 100 ml or a convenient volume of absurd water which will make full the calorimeter loving cup almost 2/three total. Since water has a density of 1 gram/cm³ = 1 gram/ml, the number of grams equals the number of ml of h2o.

5. Place ane thermometer in the banality cup and carefully work it downwardly into the metal particles.

half-dozen. When the metal is nearly finished heating, place some other thermometer into the calorimeter and record the initial temperature of the water.

7. When the metal reaches well-nigh 95°C (which is to exist the initial temperature of the metal), rapidly remove the banality cup from the boiler and pour the hot metallic into the calorimeter. Record the initial temperature of the metal.

8. Quickly put the lid on the calorimeter. Place the thermometer through the hole in the tiptop to the lid and stir the metal and water GENTLY.

nine. After you accept inserted the thermometer and stirred the mixture, record the temperature. This is the final temperature of the metallic and the water. Record the temperature. The temperature reading should drift downward slowly from this value as the mixture cools toward room temperature, so it is important to make this measurement promptly.

10. Calculate the specific heat of the metal from the information yous have collected. Make a tentative identification the the metal by referring to the table to specific heats provided.

Office Two. Heat of Fusion of H2o

Change OF Phase-SOLID TO LIQUID: An increase in internal energy is required to convert a solid to a liquid. Conversely, a reduction in internal energy can cause liquids to freeze or solidify. These solid-liquid phase changes occur without a change in temperature, i.e., no change in average kinetic energy occurs. The heat of fusion of a substance is the heat exchange required to melt one gram of the substance (calories/gm).

In this part of the experiment, the oestrus of fusion of water volition be determined. Warm water will be used to melt ice, and the alter in temperature of the water in the calorimeter will be used to compute the amount of free energy extracted from the h2o to melt the ice. The ice must absorb heat in order to cook. The oestrus absorbed can be expressed as

Heat gained past ice = Heat lost by water

(v)

where Lf is the symbol for the heat of fusion in calories/gram. The necessary oestrus will be transferred from the warm water to the ice.

PROCEDURE:

1. Warm some h2o to about 15°C higher up room temperature. (Water from the hot water tap may be hot plenty.)

ii. Mass the empty calorimeter with the styrofoam cup and record its mass. Remass it when it is approximately half full of the warm water and record the new mass. This must be washed advisedly, as it cannot exist repeated afterward. Summate the mass of the warm water.

iii. Gently stir the water and record its temperature using the 50°C thermometer.

iv. Dry pieces of the water ice provided with a piece of paper towel. Skid them into the water, stirring gently and being conscientious not to splash water out. Go along to stir the water while adding ice to it one piece at a time. When the temperature of the water is approximately fifteen°C below room temperature, end adding ice. When the final chip of water ice has melted, record the lowest temperature that the h2o reaches. This temperature is the final temperature of the warm water and of the ice h2o.

five. Mass the calorimeter and its contents and determine the mass of ice that was added.

six. Calculate the rut of fusion and find the percent deviation from the accustomed value 79.72 cal/gm.

QUESTIONS:

1. How many calories are required to heat 10 grams of h2o by xx°C?

2. What temperature change would the to a higher place amount of energy crusade in ten grams of lead?

3. How would the results for the oestrus of fusion of water have been affected if the ice were not dried before beingness placed in the calorimeter? Would this have fabricated the measured heat of fusion larger or smaller?

Specific heats (cal/gm °C)