If 6.500 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temperature of the water?

The final temperature of the water : 32.442 °C

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released  

Heat can be calculated using the formula:  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

From reaction:  

2C₆H₆ (l) + 15O₂ (g) ⟶12CO₂ (g) + 6H₂O (l) +6542 kJ, heat released by +6542 kJ to burn 2 moles of C₆H₆  

If there are 6.500 g of C₆H₆ then the number of moles:  

mol = mass: molar mass C₆H₆  

mol = 6.5 : 78  

mol C₆H₆ = 0.0833

so the heat released in combustion 0.0833 mol C₆H₆:  

the heat produced from the burning is added to 5691 g of water at 21 ∘ C  

So :  

Q = m . c . ∆T (specific heat of water = 4,186 joules / gram ° C)  

272583 = 5691 . 4.186.∆T  

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the difference between temperature and heat  

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Specific heat  

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relationships among temperature, heat, and thermal energy.  

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When heat is added to a substance  

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