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kahynickel:
A butane burner is used to heat water. The Mr of butane is 58.

standard enthalpy change of combustion of butane is –2877 kJ mol–1.
250 g of water is heated from 12 °C to 100 °C.

The burner transfers 47% of the heat released from the burning fuel to the water.

Assume that the butane undergoes complete combustion and none of the water evaporates. What is the minimum mass of butane that must be burnt?

A 0.068g          B 1.85g           C 3.94g          D 4.48g

q = mc delta T
q= 250 x 4.18x 88 = 91960 J or 91.7 KJ

2877...58 g
91.7....x g

x = 1.85 g

can someone explain ??

AWK:

kahynickel:
but what would be the effect of 47 %. the more water will be formed . but how much

47% of the heat released ..means 0.47 x 2877 = 1352.19 KJ...and then...??

Borek:

--- Quote from: kahynickel on May 18, 2019, 07:49:13 PM ---but what would be the effect of 47 %. the more water will be formed . but how much
--- End quote ---

No. 47% means not all heat will be used.

kahynickel:
1352.2 KJ will be iced up.

1352.2 KJ requires 58 g
92  KJ requires xg

x = 3.94 g.....Oh..

but why it wrote butane was completely burnt ??