Answer
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Hint: Follow the basic concepts of the calories and joules and define them by using its correlation for the required relation. Like the Energy can be expressed in various ways and the SI unit is Joule.
Complete step by step answer:
An ability to do work is known as energy. In physics, we have various types of energy Such as potential energy, kinetic energy, gravitational energy, chemical energy and many more.
A calorie is the unit of measurement. It can be expressed as the amount of heat required at a pressure of one standard atmosphere to raise the temperature of one gram of water at one degree Celsius.
Here, we are given $1{\text{ cal = 4}}{\text{.2J}}$
And $0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = ?Jk{g^{ - 1}}{K^{ - 1}}.$
Here we are asked to find the relation among the gram to kilogram and also, value for calorie is given.
Therefore, use
$
1kg = 1000gm \\
\implies 1gm = {10^{ - 3}}kg \\
$
Place the given value in the given equation-
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = \dfrac{{0.2 \times 4.2J}}{{{{10}^{ - 3}}kg.K}}$
When the exponent has negative power in the denominator then it goes to the numerator and becomes positive. Simplify the above equation –
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = \dfrac{{0.2 \times 4.2 \times 1000J}}{{kg.K}}$
Simplification implies –
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = \dfrac{{840J}}{{kg.K}}$
The above expression can be re-written as –
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = 840Jk{g^{ - 1}}{K^{ - 1}}$
Note:
Always convert the given units in the same system of units. Remember the basic conversational relations to substitute the values and convert the units. Always remember the units, derived units and SI units of the physical quantities to get the relation between the two or more physical quantities. Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
MKS System (Metre Kilogram Second)
CGS System (Centimetre Gram Second)
System International (SI)
Complete step by step answer:
An ability to do work is known as energy. In physics, we have various types of energy Such as potential energy, kinetic energy, gravitational energy, chemical energy and many more.
A calorie is the unit of measurement. It can be expressed as the amount of heat required at a pressure of one standard atmosphere to raise the temperature of one gram of water at one degree Celsius.
Here, we are given $1{\text{ cal = 4}}{\text{.2J}}$
And $0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = ?Jk{g^{ - 1}}{K^{ - 1}}.$
Here we are asked to find the relation among the gram to kilogram and also, value for calorie is given.
Therefore, use
$
1kg = 1000gm \\
\implies 1gm = {10^{ - 3}}kg \\
$
Place the given value in the given equation-
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = \dfrac{{0.2 \times 4.2J}}{{{{10}^{ - 3}}kg.K}}$
When the exponent has negative power in the denominator then it goes to the numerator and becomes positive. Simplify the above equation –
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = \dfrac{{0.2 \times 4.2 \times 1000J}}{{kg.K}}$
Simplification implies –
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = \dfrac{{840J}}{{kg.K}}$
The above expression can be re-written as –
$0.2{\text{ cal}}{{\text{g}}^{ - 1}}{}^\circ {C^{ - 1}} = 840Jk{g^{ - 1}}{K^{ - 1}}$
Note:
Always convert the given units in the same system of units. Remember the basic conversational relations to substitute the values and convert the units. Always remember the units, derived units and SI units of the physical quantities to get the relation between the two or more physical quantities. Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
MKS System (Metre Kilogram Second)
CGS System (Centimetre Gram Second)
System International (SI)
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