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Hint: Joule: Joule is SI unit of energy. If it is equal to work done by unit force (Newton) through a distance of one meter.
Calorie: Amount of energy required to raise temperature i.e. unit gram of water 60 unit degree at unit atmospheric pressure.
\[\text{1 cal=4}\text{.184 J}\]
Complete step-by-step answer:
By relation between calorie and Joule we can write
\[\Rightarrow \text{4}\text{.184 J=1 cal}\]
\[\Rightarrow \text{1 J = }\dfrac{\text{1}}{\text{4}\text{.184}}\text{cal }\]
$\Rightarrow \text{1J=0}\text{.239005cal}$
$\Rightarrow \text{1J=0}\text{.24cal}$
Hence, the answer is option c.
Note: Calories are of following types:
Small calories:- If the amount of energy required to heel unit grams to unit Celsius at unit atmospheric pressure it is generally represented in small cal.
Large calories: it is the amount of energy required to 1 kg to \[1{}^\circ C\]at 1 atm pressure. These are generally represented in caps (cal). These are also known as food calories relation between small and large.
\[\text{1 large calorie = 1 kilogram small calorie}\]
We generally consider large calories calories.
To convert \[\text{(}\mathop{E}_{j}\text{)}\]Joules to calories
\[\mathop{E}_{cal}=\dfrac{E(cal)}{4.184}calories\]
And to convert \[\text{(}\mathop{E}_{cal}\text{)}\]to \[\text{(}\mathop{E}_{j}\text{)}\]
\[\mathop{E}_{j} = 4.184\times {{E}_{cal}}\]
Calorie: Amount of energy required to raise temperature i.e. unit gram of water 60 unit degree at unit atmospheric pressure.
\[\text{1 cal=4}\text{.184 J}\]
Complete step-by-step answer:
By relation between calorie and Joule we can write
\[\Rightarrow \text{4}\text{.184 J=1 cal}\]
\[\Rightarrow \text{1 J = }\dfrac{\text{1}}{\text{4}\text{.184}}\text{cal }\]
$\Rightarrow \text{1J=0}\text{.239005cal}$
$\Rightarrow \text{1J=0}\text{.24cal}$
Hence, the answer is option c.
Note: Calories are of following types:
Small calories:- If the amount of energy required to heel unit grams to unit Celsius at unit atmospheric pressure it is generally represented in small cal.
Large calories: it is the amount of energy required to 1 kg to \[1{}^\circ C\]at 1 atm pressure. These are generally represented in caps (cal). These are also known as food calories relation between small and large.
\[\text{1 large calorie = 1 kilogram small calorie}\]
We generally consider large calories calories.
To convert \[\text{(}\mathop{E}_{j}\text{)}\]Joules to calories
\[\mathop{E}_{cal}=\dfrac{E(cal)}{4.184}calories\]
And to convert \[\text{(}\mathop{E}_{cal}\text{)}\]to \[\text{(}\mathop{E}_{j}\text{)}\]
\[\mathop{E}_{j} = 4.184\times {{E}_{cal}}\]
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