Chemistry question?
Hydrogen gives off 120. J/g of energy when burned in oxygen, and methane gives off 50. J/g under the same circumstances. If a mixture of 7.3 g of hydrogen and 10. g of methane is burned, and the heat released is transferred to 50.2 g of water at 25.0 °C, what final temperature will be reached by the water?
________°C
Hint: specific heat of H2O (l)= 4.18 J/Cxg
Answer:
H2 = 120J x 7.3 = 876 J
CH4 = 50J x 10 = 500 J
Total = 1376 J into 50.2 g water = 27.4J/g
27.4 ÷ 4.18 = 6.6°C...25 + 6.6 = 31.6°C Final temp.
So 7.3g of H2 will produce 7.3*120J of energy = 876J
10g of methane will produce 50*10J of energy = 500J
So total energy produced is 876+500 = 1376J
To raise 1 gram of H2O by 1°C takes 4.18J. So you need 50.2*4.18 J of energy to raise the temp of 50.2g of water by 1°C = 210J
Therefore 1376J will raise 50.2g of water by 6.56°C
The final water temp will be 25+6.56 = 31.56°C.
Taking into account the significant figures above, which is 2, the answer is 32°C.
NOTE: if the figures you have quoted above are not exactly as in the question the significant figures will be different. ie if 7.3 is actually 7.30 or the 10. is 10.0 then the answer should be quoted to 3 sig figures or 31.6°C
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