# How do you calculate the mass of ethanol that must be burnt to increase the temperature of 210 g of water by 65^@, if exactly half of the heat released by this combustion is lost to the surroundings?

## The heat of combustion of ethanol is $1367 k J m o {l}^{-} 1$.

Jun 1, 2018

You must burn 3.8 g of ethanol.

#### Explanation:

Step 1. Calculate the theoretical amount of heat required

The formula for the quantity of heat $q$ transferred is

color(blue)(bar(ul(|color(white)(a/a)q = mC_text(s)ΔTcolor(white)(a/a)|)))" "

where

$m \textcolor{w h i t e}{l l} =$ the mass of the object
${C}_{\textrm{s}} \textcolor{w h i t e}{l} =$ its specific heat capacity
ΔT = its change in temperature

In this problem,

$m \textcolor{w h i t e}{l l} = \text{210 g}$
${C}_{\textrm{s}} \textcolor{w h i t e}{l l} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT = "65 °C"

q = 210 color(red)(cancel(color(black)("g"))) × 4.184 "J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) ×65 color(red)(cancel(color(black)("°C"))) = "57 100"color(white)(l)"J" = "57.1 kJ"

Step 2. Calculate the actual amount of heat required

$\text{Heat required" = 57.1 color(red)(cancel(color(black)("kJ theoretical"))) × "2 kJ required"/(1 color(red)(cancel(color(black)("kJ theoretical")))) = "114 kJ}$

Step 3. Calculate the moles of ethanol required

The equation for the combustion of ethanol is

$\text{C"_2"H"_5"OH" + "⁷/₂O"_2 → "2CO"_2 + "3H"_2"O" + "1367 kJ}$

$\text{Theoretical moles" = 114 color(red)(cancel(color(black)("kJ"))) × "1 mol ethanol"/(1367 color(red)(cancel(color(black)("kJ")))) = "0.0836 mol ethanol}$

Step 4. Calculate the mass of ethanol

$\text{Mass of ethanol" = 0.0836 color(red)(cancel(color(black)("mol ethanol"))) × "46.07 g ethanol"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "3.8 g ethanol}$