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Question: When coal burns, it produces heat in the amount of \(2.5 \times {10^4} \) joules per gram. About \(4...

When coal burns, it produces heat in the amount of 2.5×1042.5 \times {10^4} joules per gram. About 40004000 joules of heat is required to raise the temperature of one kilogram of water by one degree. The amount of coal required to heat 55 kilograms of water from 10C10^\circ C to 60C60^\circ C is most nearly
(A) 1010 Grams
(B) 4040 Grams
(C) 100100 Grams
(D) 400400 Grams
(E) 16001600 Grams

Explanation

Solution

Hint Here given that we are burning coal to generate heat which is provided to water to raise its temperature by 50C50^\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 QQ . 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=mCΔTQ = mC\Delta T
where m=m = Mass
C=C = Specific heat
ΔT=\Delta T = Temperature difference

Complete Step by step answer
As we are burning coal to heat the water of 5kg5kg mass. Doing so provides heat which will be absorbed by the water which in turns will raise its temperature from 10C10^\circ C which is the initial temperature of water to its final temperature which is 60C60^\circ C . Hence temperature difference will be given by
ΔT=60C10C\Delta T = 60^\circ C - 10^\circ C
During this process, some amount of heat energy is absorbed by the water. Suppose QQ the amount of heat energy absorbed which in turn will raise the temperature of the water which is equal to ΔT\Delta T . Hence
Qwater=mwater×Cwater×ΔT{Q_{water}} = {m_{water}} \times {C_{water}} \times \Delta T -------------- Equation (1)(1)
where mwater=5kg{m_{water}} = 5kg the mass of water
Cwater=4000{C_{water}} = 4000 , Specific heat of the water
ΔT=50C\Delta T = 50^\circ C
Now substituting the values of in equation (1)(1) we get
Qwater=5kg×4000×50C\Rightarrow {Q_{water}} = 5kg \times 4000 \times 50^\circ C
Qwater=1000000J\Rightarrow {Q_{water}} = 1000000J
Qwater=106J\therefore {Q_{water}} = {10^6}J
Hence 106J{10^6}J heat energy will be required to raise the temperature of water by 50C50^\circ C .
Now it is given that 2.5×104J2.5 \times {10^4}J energy is produced by burning one gram coal. So the amount of coal required to generate the heat energy 106J{10^6}J can be given as
Amount of coal in gram mcoal=106J2.5×103J/g{m_{coal}} = \dfrac{{{{10}^6}J}}{{2.5 \times {{10}^3}J/g}}
mcoal=40g\Rightarrow {m_{coal}} = 40g
Hence 40g40g of coal is required to raise the temperature of 5kg5kg water from 10C10^\circ C to 60C60^\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 Cwater=4200{C_{water}} = 4200 which is a constant value.