How many liters of c6h14(l), measured at 20 âc, must be burned to provide enough heat to warm 27.2 m3 of water from 17.4 to 31.3 âc , assuming that all the heat of combustion is transferred to the water, which has a specific heat of 4.18 j/(gâ‹…âc)? recall that 1 ml=10â’3 l.

Given:

Volume of water = 27.2 m3

Initial temperature of water T1 = 17.4 C

Final temperature T2 = 31.3 C

Specific heat of water c  = 4.18 J/gC

To determine:

The volume (Liters) of C6H14 that must be burned to warm the given volume of water

Explanation:

Step 1: Calculate the heat required by water

Heat required to warm up the water is given as:

Q = mcΔT   (1)

m = mass of water

c = specific heat

ΔT = change in temp of water = T2-T1

Now,

Density of water = 1 g/cm3

Volume of water = 27.2 m3 = 27.2 * 10⁶ cm3

Thus, mass of water = Density * volume = 27.2 * 10⁶ g

Substituting the values in equation (1) we get

Q = 27.2*10⁶g * 4.18 J/gC * (31.3-17.4)C = 1.580*10⁹ J

Step 2: Calculate the moles of C6H14 that need to be burned

Heat of combustion of C6H14 = 4163 * 10³ J/mol

# moles corresponding to 1.580*10⁹ J  is given as-

= 1.580 * 10⁹/4163*10³ = 379.53 moles

Step 3: Calculate the volume of C6H14 required

Now, molar mass of C6H14 = 86 g/mol

Mass of C6H14 corresponding to the calculated moles = 379.53 * 86

= 3.264*10⁴ g

Density of C6H14 = 0.6606 g/ml

Volume of C6H14 = Mass/Density = 32640 g/0.6606 g.ml-1 = 49409 ml

Ans: Volume of C6H14 required = 49.4 L

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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