SUMMARY OF THE Theory
Intilah heat was first introduced by Antoine Laurent Lavoisier intheyear 1743 Until 1794, the French.
The theory presented Lavoisier much opposed, AmongthemBenjamin Thomson or Count Rumford (1753 - 1814) ofAmerica.
Later That Prescott Joule heat is a form of energy. From the resultsof experiments Joule concluded That the comparison Between thework done is a fixed number, the which the 4186 rounded to 4200.
Based on the conclusion That heat is a form of energy, then theunitto heat the same as the energy unit.
So the definition of heat is: a form of energy fromhigher Thatmoves the object to the object temperature, the temperature islower Nowhere Pls two objects touch.
Based on the heat needed to raise the temperature of an object /substance Depends on three factors, namely:
1. mass substances.
2. Kinds of substances
3. temperature rise of the substance.
Relations with the heat density of a substance That Is Thatthegreater mass of substance, the Greater the heat energy neededtoraise its temperature.
Magnitude relationship of heat to the type of substance.
From the experimental results cans be concluded That thesubstanceof heat Necessary to raise its temperature Depends onthe type ofsubstance.
Relationships heat with temperature rise.
From the results of experiments it was found That for asubstancewith the Same type and mass, Pls heated with adifferent amount of heat will from Produce a different temperaturerise. Thus, the Greater the heat That is given to an object, theGreater the increase is intemperature.
After conducting experiments installments cans be obtained fromthe relationship of heat with the mass density, type of substance,andsubstance temperature rise. Mathematically, the relationshipisformulated as follows:
• Q = m. c. Δt
where:
Diperlukaan Q = heat (Joule)
m = mass density (kg)
c = heat type (Joule / kg ⁰ C or Joule / kg ⁰ K)
Δt = temperature rise (⁰ ⁰ C or K)
SAMPLE QUESTIONS:
1. Who first used intilah heat?
replied: Antoine Laurent Lavoisier.
2. What is the definition of heat?
answer: Heat is a form of energy that moves from highertemperature object to the object where the temperature is lowerwhen two objects touch.
3. There are three factors that affect heat to raise the temperatureof an object, say?
answer: the mass of matter, kinds of substances, andtemperature rise substance.
Intilah heat was first introduced by Antoine Laurent Lavoisier intheyear 1743 Until 1794, the French.
The theory presented Lavoisier much opposed, AmongthemBenjamin Thomson or Count Rumford (1753 - 1814) ofAmerica.
Later That Prescott Joule heat is a form of energy. From the resultsof experiments Joule concluded That the comparison Between thework done is a fixed number, the which the 4186 rounded to 4200.
Based on the conclusion That heat is a form of energy, then theunitto heat the same as the energy unit.
So the definition of heat is: a form of energy fromhigher Thatmoves the object to the object temperature, the temperature islower Nowhere Pls two objects touch.
Based on the heat needed to raise the temperature of an object /substance Depends on three factors, namely:
1. mass substances.
2. Kinds of substances
3. temperature rise of the substance.
Relations with the heat density of a substance That Is Thatthegreater mass of substance, the Greater the heat energy neededtoraise its temperature.
Magnitude relationship of heat to the type of substance.
From the experimental results cans be concluded That thesubstanceof heat Necessary to raise its temperature Depends onthe type ofsubstance.
Relationships heat with temperature rise.
From the results of experiments it was found That for asubstancewith the Same type and mass, Pls heated with adifferent amount of heat will from Produce a different temperaturerise. Thus, the Greater the heat That is given to an object, theGreater the increase is intemperature.
After conducting experiments installments cans be obtained fromthe relationship of heat with the mass density, type of substance,andsubstance temperature rise. Mathematically, the relationshipisformulated as follows:
• Q = m. c. Δt
where:
Diperlukaan Q = heat (Joule)
m = mass density (kg)
c = heat type (Joule / kg ⁰ C or Joule / kg ⁰ K)
Δt = temperature rise (⁰ ⁰ C or K)
SAMPLE QUESTIONS:
1. Who first used intilah heat?
replied: Antoine Laurent Lavoisier.
2. What is the definition of heat?
answer: Heat is a form of energy that moves from highertemperature object to the object where the temperature is lowerwhen two objects touch.
3. There are three factors that affect heat to raise the temperatureof an object, say?
answer: the mass of matter, kinds of substances, andtemperature rise substance.
4. Minyak 500gr land mass which is heated so that its temperaturerose 10 ° C. If the heat of the kerosene type 2.2 x 10 ³ J / kg ° C,calculate the amount of heat required to heat the oil?
Given: m = 500gr = 0.5 kg
Cminyak soil = 2.2 x 10 ³ J / kg ° C
Δt = 10 ° C
Asked: Q = ......?
Given: m = 500gr = 0.5 kg
Cminyak soil = 2.2 x 10 ³ J / kg ° C
Δt = 10 ° C
Asked: Q = ......?
• Q = m x c x Δt
= 0.5 x 2.2 x 10 ³ x 10
= 11 x 10 ³ joule
= 11 000 joules
= 11,000 x 0.24 calories
= 0.5 x 2.2 x 10 ³ x 10
= 11 x 10 ³ joule
= 11 000 joules
= 11,000 x 0.24 calories
5. A kettle which is made of aluminum with a mass of 0.5 kg of initialtemperature 10 ° C, then given a heat of 45 000 J.
(calor type aluminum = 900 J / kg ° C)
What is the temperature of the kettle now?
Given: m = 0.5 kg; t1 = 10 ° C
Q = 45,000 Joules
c = 900 J / kg ° C
Asked: t2 = .....?
(calor type aluminum = 900 J / kg ° C)
What is the temperature of the kettle now?
Given: m = 0.5 kg; t1 = 10 ° C
Q = 45,000 Joules
c = 900 J / kg ° C
Asked: t2 = .....?
Q = m x c x Δt
Δt = Q / (m x c)
= 45,000 / (0.5 x 900)
= 100 ° C
Δt = t2 - t1
t2 = t1 + Δt
= 100 + 10
t2 = 110 ° C
Δt = Q / (m x c)
= 45,000 / (0.5 x 900)
= 100 ° C
Δt = t2 - t1
t2 = t1 + Δt
= 100 + 10
t2 = 110 ° C
6. How much heat required to raise the temperature of 30 ° C copper.If the mass of copper = 500 grams and copper heat 0.09 cal / g ° C?
Given: Δt = 30 ° C; m = 500gr
c = 0.09 cal / g ° C
Asked Q = .......?
Given: Δt = 30 ° C; m = 500gr
c = 0.09 cal / g ° C
Asked Q = .......?
Answer:
Q = m x c x Δt
= 500 x 0.09 x 30
= 1350 calories
Q = m x c x Δt
= 500 x 0.09 x 30
= 1350 calories
7. To raise the temperature of 500 g of copper from 10 ° C to 110 °C required 20 000 Joule
heat.
What kind of copper heat?
Given: m = 500gr
Δt = t2 - t1 = 110-10 = 100 ° C
Q = 20,000 Joules
Q = m x c x Δt
20 000 = 500 x c x 100
c = 20 000 / (500x100)
= 20000/50000
= 0.4 Joule / g ° C
If the unit converted to cal / g ° C, then:
c = 0.4 x 0.24 cal / g ° C
= 0.096 cal / g °
20 000 = 500 x c x 100
c = 20 000 / (500x100)
= 20000/50000
= 0.4 Joule / g ° C
If the unit converted to cal / g ° C, then:
c = 0.4 x 0.24 cal / g ° C
= 0.096 cal / g °
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