# Calculate the amount of heat required to convert 45.0 g of water at 43.0 ∘C to steam at 100.0 ∘C.?

## Express your answer to three significant figures and include the appropriate units.

Apr 6, 2018

$\approx 112.6 k J$

#### Explanation:

First, the water must be heated to 100 degrees celcius and then the 100 degree water must be turned into steam.

Using the equation $Q = m c \Delta T$ we can calculate the amount of energy for heating the water to 100 degrees.

$Q$=energy input
$m$=mass of the matter to heat
$c$=specific heat capacity of the matter to heat
$\Delta T$=the temperature change of the matter

We know
m=45 g
c=4187 Joules per kilogram- the specific heat capacity of water.
$\Delta T$ = 100-43=57, how many degrees we must increase the water by to make it boil and turn into steam.

Hence we find $Q$
$Q = \left(0.045\right) \times \left(4187\right) \times \left(57\right)$
$Q = 10721 \approx 10.7 k J$

Now we must turn the water into steam. This is done by increasing the potential energy of the water without changing its temperature using something called Specific latent heat of vaporisation of water- which is the energy amount required to convert a a unit mass of water into steam at constant temperature.

It is given by the equation
$Q = m L$
$Q$= energy input
$m$=mass of the matter to vaporize
$L$=the specific latent heat

According to Wikipedia the value of the latent heat of vaporisation of water is:
$L = 2264.705$ kj/kg

Plugging it in to the equation:

$Q = 2264.705 \times 0.045 \approx 101.9 k J$

So the total energy change from both of these coverstion are:
$101.9 + 10.7 \approx 112.6 k J$

$112.6 k J$