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**Hint**Here given that we are burning coal to generate heat which is provided to water to raise its temperature by $50^\circ C $. Hence to find how much heat energy is required to raise the temperature of the water we will use the heat energy equation $Q $. Now comparing this evaluated amount of heat with the provided heat in the question we will get the amount of coal required for this process.

Formula used:

Heat energy equation

$Q = mC\Delta T $

where $m = $Mass

$C = $Specific heat

$\Delta T = $Temperature difference

**Complete Step by step answer**

As we are burning coal to heat the water of $5kg $ mass. Doing so provides heat which will be absorbed by the water which in turns will raise its temperature from $10^\circ C $ which is the initial temperature of water to its final temperature which is $60^\circ C $. Hence temperature difference will be given by

$\Delta T = 60^\circ C - 10^\circ C $

During this process, some amount of heat energy is absorbed by the water. Suppose $Q $ the amount of heat energy absorbed which in turn will raise the temperature of the water which is equal to $\Delta T $. Hence

${Q_{water}} = {m_{water}} \times {C_{water}} \times \Delta T $ -------------- Equation $(1) $

where ${m_{water}} = 5kg $the mass of water

${C_{water}} = 4000 $, Specific heat of the water

$\Delta T = 50^\circ C $

Now substituting the values of in equation $(1) $we get

$ \Rightarrow {Q_{water}} = 5kg \times 4000 \times 50^\circ C $

$ \Rightarrow {Q_{water}} = 1000000J $

$\therefore {Q_{water}} = {10^6}J $

Hence ${10^6}J $ heat energy will be required to raise the temperature of water by $50^\circ C $.

Now it is given that $2.5 \times {10^4}J $energy is produced by burning one gram coal. So the amount of coal required to generate the heat energy ${10^6}J $can be given as

Amount of coal in gram ${m_{coal}} = \dfrac{{{{10}^6}J}}{{2.5 \times {{10}^3}J/g}} $

$ \Rightarrow {m_{coal}} = 40g $

Hence $40g $ of coal is required to raise the temperature of $5kg $water from $10^\circ C $ to $60^\circ C $.

**Note**While solving the numerical one should ensure each physical quantity is in its SI units. If not then proceed with converting it first. Here the specific heat of water is given but if it is not provided that one should take the specific heat of the water as ${C_{water}} = 4200 $ which is a constant value.

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