Question

How many grams of c5h12 must be burned to heat 1.38 kg of water from 20.4 °c to 98.8 °c? assume that all the heat released during combustion is used to heat the water.

Answer 1

The mass (in grams) of C₅H₁₂ that must be burned to heat 1.38 kg of water from 20.4 °C to 98.8 °C, is 9.29 g

How to calculate the mass of C₅H₁₂ that must be burned?

First, we shall calculate the heat absorbed by the water to raise its temperature from 20.4 °C to 98.8 °C. Details below:

• Mass of water heated (M) = 1.38 Kg
• Initial temperature of water (T₁) = 20.4 °C
• Final temperature of water (T₂) = 98.8 °C
• Temperature change of water (ΔT) = 98.8 °C - 20.4 = 78.4 °C
• Specific heat capacity of water (C) = 4184 J/Kg°C
• Heat absorbed (Q) =?

Q = MCΔT

= 1.38 × 4184 × 78.4

= 452675.328 J

Next, we shall the mole of pentane, C₅H₁₂. Details below:

• Heat absorbed by water (Q) = 452675.328 J
• Heat released by pentane (Q) = -452675.328 J
• Heat of combustion of pentane (ΔH) = -3509 kJ/mol = -3509 × 1000 = -3509000 J/mol
• Mole of pentane (n) =?

`n = (Q)/(\Delta H) \\\\n = (-452675.328)/(-3509000) \\\\n = 0.129\ mole`

Finally, we shall calculate the mass of pentane, C₅H₁₂ that must be burned. This is shown below:

• Mole of pentane = 0.129 mole
• Molar mass of pentane = 72 g/mol
• Mass of pentane =?

Mass of pentane = Mole × molar mass

Mass of pentane = 0.129 × 72

= 9.29 g

Answer 2

The mass of the compound that will produce the heat is 9.3 g.

We have that the heat evolved is;

H = mcdT

H = heat evolved

c = heat capacity of water

dT = temperature change

H = 1.38 * 4200 * (98.8 - 20.4)

= 454.4 kJ

The heat of reaction can be gotten from the thermochemical equation;

`C_5H_12}(l) + 8O_2(g) ---- > 5CO_2(g) + 6H_2O(l)` ΔH°rxn= –3,510 kJ/mol

If 1 mole of the compound produces 3,510 kJ

x moles of the compound produces 454.4 kJ

x = 0.129 moles

Mass of the compounds = 0.129 moles * 72 g/mol

= 9.3 g

The mass is 9.3 g