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**Hint:**This question is based on the principle of calorimetry which tells us total energy given by the hot body is always equal to the total energy taken by the cold body. Also the concept of electrical power & corresponding concept of units consumed is involved.

**Formula used:**

$\left( 1 \right)\;$ ${{\Delta Q = ms\Delta T}}$

Where, ${{\Delta Q = }}$ amount of heat given/ heat taken in ${{J}}{{.}}$

${{m = }}$ mass of the body in $kg$

$s = $ specific heat capacity ${{KJ/Kg}}\;{\;^{{o}}}{{C}}$

$\left( 2 \right)$ ${{1}}\;{{unit}}\;{{ = 1KWh}}\;{{ = 3}}{{.6 \times 1}}{{{0}}^{{6}}}{{J}}$

$\left( 3 \right)$ Specific heat capacity of water ${{ = }}\;{{4}}{{.2}}\;{{J/g}}{\;^{{o}}}{{C}}$

**Complete step by step solution:**

(a) Given $1$ cup of tea contains ${{200cc}}$ of water, so, volume of the water boiled in $4$ cups

${{ = }}\;{{4 \times 200cc}}$

${{ = 800cc}}$

Total mass of the water to be boiled $ = $ Total volume of the water be boiled $ \times $ Density of water since Density of water is ${{1gm/c}}{{{m}}^{{3}}}$; therefore mass of water to be boiled ${{ = }}\;{{800c}}{{{m}}^{{3}}}{{ \times 1gm/c}}{{{m}}^{{3}}}{{ = 800g}}$

As ${{100g}}$ ${{ = }}\;{{1KG}}$

Mass of water to be boiled is equals to $\dfrac{{800}}{{1000}} = 0.8{{kg}}$

It is also given that the initial temperature of water is ${25^ \circ }{{C}}$.

The temperature of water $ = \;$ Boiling point of water $ = {100^ \circ }{{C}}$

$\therefore $ Rise in temperature of water ${{ = }}\;{{10}}{{{0}}^{{o}}}{{C - 2}}{{{5}}^{{o}}}{{C}}\;{{ = 7}}{{{5}}^{{o}}}{{C}}$

We know that the specific heat capacity of water

\[{{S = 1}}\;{{cal/}}{{{g}}^{{o}}}{{C}}\;{{ = 4}}{{.2J/}}{{{g}}^{{o}}}{{C}}\]

$ \Rightarrow {{S = 4200J/K}}{{{g}}^{{o}}}{{C}}$

Now, ${{\Delta Q}}\left( {{{heat}}\;{{required}}} \right)\;{{ = ms\Delta T}}$

${{m = }}$ mass of water in kg

${{s = }}$ specific heat of water

${{\Delta T = }}$ change in temperature

Putting the values

${{\Delta Q}}\;{{ = 0}}{{.8 \times 4200 \times 75}}\;{{ = 252000J}}$

We know that ${{1}}\;{{unit}}\;{{ = }}\;{{10000Whr}}\;{{ = 3}}{{.6 \times 1}}{{{0}}^{{6}}}{{J}}$

Now, heat required in terms of units $ = $

$H = \dfrac{{{{252000}}}}{{{{3}}{{.6 \times 1}}{{{0}}^{{6}}}}}{{ = 0}}{{.07}}\;{{units}}$

As given in the question Rs $1$ is $1$ unit, so the total cost to heat $4$ cups of water from

${25^ \circ }{{C}} \to {{0}}{{.07Rs = 7}}\;{{paise}}$

(b) Here we are given that the initial temperature of the water is ${5^ \circ }{{C}}{{.}}$

And we know that final temperature of water $ = {100^ \circ }{{C}}$

(Boiling point of water)

$\therefore $ Rise in temperature of water $ = {100^ \circ } - {5^ \circ }\; = \;{95^ \circ }{{C}}$

Now, ${{\Delta Q}}\;\left( {{{heat}}\;{{required}}} \right)\;{{ = }}\;\;{{ms\Delta T}}$

$

{{m = }}\;{{mass}}\;{{of}}\;{{water}} \\

{{s}}\;{{ = }}\;{{specific}}\;{{heat}}\;{{of}}\;{{water}} \\

{{\Delta T}}\;{{ = }}\;{{change}}\;{{in}}\;{{temperature}} \\

$

Putting the values,

${{\Delta Q}}\;{{ = 0}}{{.8 \times 4200 \times 95}}\;{{ = }}\;\;{{319,200J}}$

Now, heat required in terms of units $ = $

$\dfrac{{{{319200}}}}{{{{3}}{{.6 \times 1}}{{{0}}^{{6}}}}}\;{{ = 0}}{{.0886J}}$

$ \approx 0.09\;{{units}}$

So the cost of boiling $4$ cups of water if the room temperature drops to

${5^ \circ }{{C}}\; \to {{0}}{{.09}} \times \;1$

${5^ \circ } \to 0.09$

${{{5}}^ \circ }{{C}} \to {{9 paise}}$

**Additional Information:**

In this question the specific heat capacity of water, density of water are to be remembered. This data is not given in the question cannot be solved. Also the boiling point of water is ${100^ \circ }{{C}}$ is to be remembered.

**Note:**The specific heat capacity of water is defined as the amount of heat required for unit mass of water to raise its temperature by ${1^ \circ }{{C}}{{.}}$ Its value ${{ = }}\;{{4}}{{.2J/}}{{{g}}^{{o}}}{{C}}.$

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