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3 years ago

Butane, C4H10 has a standard enthalpy of combustion= -2881 kJ/mol

Chemistry

1 answer:

Bad White [126]3 years ago

5 0

Answer:

14.91 K.

Explanation:

To solve this problem, we can use the following relation:

Q = m.c.ΔT,

where, Q is the amount of heat transferred to water.

m is the mass of the amount of water (m = 2.0 kg = 2000.0 g).

c is the specific heat capacity of water (c = 4.2 J/g.K).

ΔT is the change in temperature due to the transfer of butane burning.

To determine Q that to be used in calculation:

Q from 4.000 g of butane is completely burned is - 198.3 kJ = 198300 J.

The negative sign symbolizes the the enthalpy change is exothermic, which means that the energy is released.

Note that only 63.15% of the energy generated is actually transferred to the water.

∴ Q (the amount of heat transferred to water) = (198300 J)(0.6315) = 125226.45 J.

Now, we can obtain the change in temperature:

∴ ΔT = Q/m.c. = (125226.45 J) / (2000.0 g)(4.2 J/g.K) = 14.9079 K ≅ 14.91 K.

This means that the temperature is increased by 14.91 K.

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