Answer to Question #119309 in Organic Chemistry for emily

Question #119309
If 3.65g of butane is burned underneath a cup holding 1.00 L of water at 21.0o C, what will the final temperature of the water be (ΔHcomb = -3325 kJ/mol)? (3 marks)
1
Expert's answer
2020-06-04T10:32:10-0400

The combustion reaction for butane is:


2C4H10 + 13O2 = 8CO2 + 10H2O + 3325 kJ/mol


We have 3.65/58 = 0.0629 mol of butane, so during combustion this amount of butane will be released 3325*0.0629 = 209.2 kJ of heat.


Q = c*m*Δt, where Q - amount of heat, C - heat capacity for wather (4.19 kJ/kg*K), m - mass of water (we have 1.00 L = 1.00 kg).


Δt = Q/(c*m) = 209.2 / (1*4.19) = 49.94 oC


The final temperature of the water be 21.00 + 49.94 = 70.94 oC



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