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How many calories of heat are required by grams of water at $99^\circ {\text{C}}$ to boil off?
(A) $530$
(B) $640$
(C) $540$
(D) $500$

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Last updated date: 16th Jun 2024
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Answer
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Hint: In this question, the first the heat is required to boil the water from $99^\circ {\text{C}}$ to $100^\circ {\text{C}}$ and then the heat required to convert the boiled water to steam at $100^\circ {\text{C}}$. After that the total heat required is needed to convert from Joule to Calorie.

Complete step by step answer:

In the above given question, we have to calculate the amount of heat required to boil $1\;{\text{gm}}$ of water at $99^\circ C$.

We know that, first the amount of heat required to boil the water from $99^\circ {\text{C}}$ to $100^\circ {\text{C}}$, then the heat required to convert the boiled water to steam at $100^\circ {\text{C}}$.

As we know that the latent heat of vaporization is the amount of heat required to convert the unit mass of boiled water to the steam at constant temperature. The latent heat of vaporization for water is $2260$ J/g

In this question, first the amount of heat required to boil the water from $99^\circ {\text{C}}$ to $100^\circ {\text{C}}$, then the heat required to convert the boiled water to steam at $100^\circ {\text{C}}$.

So, the total amount of heat is required is calculated as,
$Q = m{c_p}\left( {{T_2} - {T_1}} \right) + mL$

Here, the specific heat at constant pressure of the water is ${c_p} = 4.2$ J/g$^\circ$C, the mass of the water is $m$, and the latent heat of vaporization of water is $L$.

Now, substitute the given values in the above expression as,

$
  Q = \left( 1 \right)\left( {4.2} \right)\left( {100 - 99} \right) + \left( 1 \right)\left( {2260} \right) \\
  \Rightarrow Q = 4.2 + 2260 \\
  \Rightarrow Q = 2264.2\;{\text{J}} \\
 $

Now, convert the amount of heat from Joule to calorie as we know that $1\;{\text{cal}} = 4.2\;{\text{J}}$,so
$
  Q = \dfrac{{2264.2}}{{4.2}}\;{\text{cal}} \\
  Q = 539.09\;{\text{cal}} \\
  Q \approx 540\;{\text{cal}} \\
 $

Hence, the correct option is C.

Note:
The definition of calorie is the amount of heat required to raise the temperature of $1\;{\text{g}}$ of water to $1^\circ {\text{C}}$. One should also remember the value of specific heat of water in order to solve the problems. Water molecules interaction drives the boiling points and freezing. Considerable amount of energy is required to break water molecules, so there is not much difference in temperature of water on heating.