# Find the temperature increase in ""^@ "C" expected for "1.00 L" of water when it absorbs all of the energy from the combustion of "1.00 g" of acetylene, "C"_2"H"_2(g)? (Every mole of acetylene releases "1300 kJ" of energy when combusted.)

## a) ${11.8}^{\circ} \text{C}$ b) ${14.1}^{\circ} \text{C}$ c) ${15.5}^{\circ} \text{C}$ d) ${13.4}^{\circ} \text{C}$ e) ${13.6}^{\circ} \text{C}$

Jan 25, 2018
1. 11.8

#### Explanation:

Molecular weight of acetylene is (12*2+2*1) or $26 g m$

So, $26 g m$ of acetylene contain $1$ mol,hence $1 g m$ contain $\frac{1}{26}$ mole

Now, if per mole acetylene releases $1300 K J$ of energy then,for $\frac{1}{26}$ moles energy released will be $\left(\frac{1300 \cdot {10}^{3}}{4.2}\right) \cdot \left(\frac{1}{26}\right)$ Calorie.

Now,this amount of heat($H$) will be taken up by $1 l i t e r$ of water,

$1 l i t e r = 1000 c {m}^{3}$ and density of water is $1 \frac{g m}{c {m}^{3}}$

So,mass of $1 l i t e r$ of water is $1000 g m$(as, mass = volume*density)

So, we can apply $H = m s d \theta$ (where $m$ is the mass of water,$s$ is specific heat of water and its value is 1 CGS unit, and $d \theta$ is the change in temperature)

Putting the values we get, $d \theta$ = $11.9$ degree C