# How to calculate the specific heat of ethanol?

100 g of water was heated from $\pu{15 °C}$ to $\pu{70 °C}$ by burning $\pu{20 g}$ of ethanol. How much energy per mole does this ethanol have?

So far I have calculated that there are $\pu{0.43 mol}$ in $\pu{20 g}$ of ethanol using the formula $n = m/M$. \begin{align} m &= \pu{20 g}\\ M &= 12 + 3 + 12 + 2 + 16 + 1 = \pu{46 g/mol}\\ n &= 20/46 = \pu{0.43 mol}\\ \end{align}

From here, I'm not sure if I should divide the energy used to heat up the water by the amount of substance. By using the specific heat formula,

$$Q = c \cdot m \cdot \Delta{}T = \pu{4.2 J//g \cdot K} \cdot \pu{100 g} \cdot \pu{55 K} = \pu{23100 J}$$

$$\text{Energy per mole} = \frac{\pu{23100 J}}{\pu{0.43 mol}} = \pu{53720.93 J//mol}.$$

Each gram of water has an specific heat of 4 J/(K.g)*. So heating $\pu{100 g}$ of water from $\pu{15 °C}$ to $\pu{70 °C}$ would take:

Heat = (70°C - 15°C) * 100g * 4 J/(K.g) = 22,000 joules = 22 kJ


$\pu{20 g}$ of ethanol would be:

Ethanol mol = (20g / 46,07 g/mol) = 0.43 mol


So this means that:

0.43 mol = 22 kJ
mol = 22 / 0.43
mol = 50677 J = 50.67 kJ


So ethanol has 50.67 kJ/mol vs 53.72 kJ/mol said by you. I would say that you have made fine your homework, there is a very little difference (I have used some decimal to calculate, maybe you not. (Below you can find a more accurate answer)).

* Water specific heat: water changes it's specific heat when it changes it's heat. For example: (Wikipedia)

Water at  100 °C (steam)  2.080  J/(K.g)
Water at   25 °C          4.1813 J/(K.g)
Water at  100 °C          4.1813 J/(K.g)
Water at  −10 °C (ice)    2.05   J/(K.g)

Normaly it's used:        4.1813 J/(K.g)


A better table can be found in a french page.
Using this table we can do a better aproximation:

Water heat = (70°C - 15°C)        * 100g
Water heat = (293.03 J - 63.04 J) * 100g
Water heat =    229.99 J          * 100g
Water heat = 22,999 J = 22.999 kJ ≈ 22.9 kJ

Ethanol mol = (20g / 46.07 g/mol) = 0.43 mol
Exactly: 0,4341219882787063164749294551769 mol

0.43 mol = 22999 J
mol = 22999 / 0.43
mol = 52978.19 J = 52.97819 kJ ≈ 52.97 kJ
Exactly: 52978,196500000000000000000000001 J

Answer: Ethanol has 52.97 kJ/mol. A value closer to your own value.


PD: I am only a normal person who like chemistry, maybe I am wrong. For example in the theory I have no idea what means $Q = c \cdot m \cdot \Delta{}T = \pu{4.2 \frac{J}{g \cdot K}} \cdot \pu{100 g} \cdot \pu{55 K} = \pu{23100 J}$ Well, I have some ideas but I haven't learned it in the school.

# Edit:

I notice that you have your own mol of ethanol (46 g/mol instead of my 46.07 g/mol) and your own specific heat of water (4.2 J/(K.g)). You have to do:

Water heat = (70°C - 15°C) * 100g * 4.2 J/(K.g)
Water heat =      55°C     * 100g * 4.2 J/(K.g)
Water heat =
Water heat = 23100 J = 23.1 kJ

Ethanol mol = (20g / 46 g/mol) = 0.43 mol
Exactly: 0.43478260869565217391304347826087 mol

0.43 mol = 23100 J
mol = 23100 / 0.43
Using mol decimals (0.43478260869565217391304347826087)
mol = 53130 k = 53.13 kJ
Using only 2 decimals (0.43)
mol = 53720,93 J ≈ 53.72 kJ
Exactly: 53720,930232558139534883720930233 J

Answer: Ethanol has 53.13 kJ/mol (With decimals) or 53.72 kJ (with 2 decimal). So yes! You have done right!

• Kimchiboy03 assumed a heat capacity of $\pu{0.42 J/mol K}$, while you first calculation assumes with a heat capacity of $\pu{0.4 J/mol K}$ a value that is almost $\pu(5%}$ smaller than the former. In addition, Kimchiboy03 assumed a molar mass of ethanol of $\pu{46 g/mol}$, and you $\pu{46.07 g/mol}$. These contribute to numerical differences in the results. Eventually, I would round $\pu{52978 J}$ up to $\pu{52.98 kJ/mol}$, rather than down to $\pu{52.97 kJ/mol}$. Even if this may be seen as nit-picking. – Buttonwood Jun 17 '17 at 0:02
• @Buttonwood, oh, I used the value of the mol from Wikipedia. – Ender Look Jun 17 '17 at 2:21