Question

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.

Answer 1

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

Step-by-step 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