# How much energy is required to convert 429 g of water to steam at 156 °C?

## ${C}_{\textrm{w a t e r}} = \text{4.184 J/(°C·mol)"; C_text(vapour) = "2.078 J/(°C·mol)"; Δ_text(vap)H = "2257 J/g}$

Sep 12, 2017

$\approx 1155 \frac{k J}{g {\cdot}^{o} C}$.

#### Explanation:

We will be using the formula: $m c \Delta T$, where:
=> $m$ is the mass of the object in grams.
=> $c$ is the specific heat capacity of the object.
=> $\Delta T$ is the change in temperature found by subtracting the initial temperature from the final.

Let's go over what we are given:

$m = 429 \text{g}$

${c}_{\text{liquid water}} = 4.184 \frac{J}{g {\cdot}^{o} C}$

${c}_{\text{water vapour}} = 2.078 \frac{J}{g {\cdot}^{o} C}$

$\Delta T = {T}_{\text{final" - T_"initial}}$

Now there's something we have to consider, we can't just get from liquid at ${24}^{o} C$ to steam at ${156}^{o} C$. What we have to do is convert from liquid to it's boiling point of ${100}^{o} C$, and then from ${100}^{o} C$ to ${156}^{o} C$.

So let's do the liquid to gas first.

${q}_{\text{liquid water}} = m c \Delta T$

$= \left(429\right) \left(4.184\right) \left(100 - 24\right)$

$= 136415.136 \frac{J}{g {\cdot}^{o} C}$

Next, we convert the liquid at 100 °"C" to vapour at 100 °"C". For this, we

${q}_{\text{conversion}} = m \Delta H$

$= 429 \left(2257\right)$

$= 968253$

Now let's do the gas to hotter gas.

${q}_{\text{water vapour}} = m c \Delta T$

$= \left(429\right) \left(2.078\right) \left(156 - 100\right)$

$= 49921.87 \frac{J}{g {\cdot}^{o} C}$

Now all we have to do is add the energies.

${q}_{\text{total"=q_"liquid water" + q_"conversion" + q_"water vapour}}$

$= 136415.136 + 968253 + 49921.87$

$= 1154590.01 \frac{J}{g {\cdot}^{o} C}$

$= 1154.59001 \frac{k J}{g {\cdot}^{o} C}$

$\approx 1155 \frac{k J}{g {\cdot}^{o} C}$

Hope this helps :)