0.46g of ethanol when burned raised the temperature of 50g of water by 14.3K. Calculate the heat of combustion of ethanol.
[C = 12, O = 16, H = 1, Specific heat capacity of water = 4.2 jg-1K-1]
+3000 KJ mol-1
+300 KJ mol-1
-300 KJ mol-1
-3000 KJ mol-1
Explanation
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H=-mcโT
H=-50ร4.2ร14.3
H=-3003
0.46/46=0.01
H=-3003/0.01
H=-300300J/mol
H=-3003JK/mole
Mass of water =50g. Change in temp= 14.3k. Q-mcโt. Q-50ร4.2ร14.3= -3003j. 0.46g of ethanol liberate -3003j. 46g of ethanol will liberate=46ร-3003/0.46. =-300300j/mol ~ -300.3kj/mol
No admin from your calculations I have 3003.
But guys check this out and correct me if I'm wrong.
First find the number of moles used by using the formula n=m/M where m= mass and M=molar mass
n=0.46/46
=0.01mole
๏ผพ๏ผจ๏ผ๏ผmc๏ผพt Note: the sign ^ is used as change in .
^H=-(0.05kg*4.2*14.3)
^H=-3.003
Finally you divide your answer by "n"
=^H/n
=-3.003/0.01
=-300.3
=-300kj/mol
what the question is asking for is that, 0.4g of ethanol react with the heat of combstion of water, then 46g which is tge molecular mass if ethanol will react with x gram.
โH= -mcโT
= -3003J/K :. -3.003KJ/K
0.46 = -3.003
46 = X.... X = -300.3KJ/K//
1). The question asked for the heat of combustion of ethanol, why was the parameters of water used instead
2). okay, let's agree that's the correct way to solve it, but the answer is wrong...(-50*4.2*14.3 = 3003) but the answer chosen was -300KJ when it should be -3003KJ
Please look into it MYSCHOOL
thank you
