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Assumptions: the specific heat capacity of water is $\rm 4.182\; J \cdot mol^{-1}$ , the melting point of water is $\rm 0\, ^{\circ} C$ , and that the boiling point of water is $\rm 100 \,^{\circ} C$ .

Explanation:

It takes five steps to convert 13.0 grams of $\rm \text{-}10.0\, ^{\circ}C$ ice to steam at $\rm 111.0\,^{\circ}C$ .

Step one: heat the 13.0 gram of ice from $\rm \text{-}10.0\, ^{\circ}C$ to $\rm 0\,^{\circ}C$ . The change in temperature would be $\rm 10.0\,^{\circ}C$ .
Step two: supply the heat of fusion to convert that 13.0 gram of ice to water.
Step three: heat the 13.0 gram of water from $\rm 0\,^{\circ}C$ to $\rm 100\,^{\circ}C$ . The change in temperature would be $\rm 100\,^{\circ}C$ .
Step four: supply the heat of vaporization to convert that 13.0 gram of water to steam.
Step five: heat the 13.0 gram of steam from $\rm 100\,^{\circ}C$ to $\rm 111.0\,^{\circ}C$ . The change in temperature would be $\rm 11.0\,^{\circ}C$ .

<h3>Energy required for step one, three, and five</h3>

The following equation gives the amount of energy $Q$ required to raise the temperature of an object by a $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

In this equation,

$c$ is the specific heat of this substance,
$m$ is the mass of the substance, and
$\Delta T$ is the change in the temperature of the object.

Assume that there's no mass loss in this whole process. The value of $m$ would stay the same at $13.0\; \rm g$ .

$\begin{aligned}& &&\text{Energy required for raising temperature} \cr &=&& c(\text{Ice}) \cdot m \cdot \Delta(\text{Ice}) \cr & && + c(\text{Water}) \cdot m \cdot \Delta(\text{Water})\cr & && + c(\text{Steam}) \cdot m \cdot \Delta(\text{Steam}) \cr & = && (2.09 \times 13.0 \times 10) \cr & && + (4.182 \times 13.0 \times 100) \cr & &&+ ( 2.01 \times 13.0 \times 10) \cr & = && 5969.6\;\rm J \cr & = && 5.969\; \rm kJ\end{aligned}$ .

<h3>Energy required for step two and four</h3>

The equations for the energy of fusion and energy of vaporization are quite similar:

$E(\text{Fusion}) = n \cdot \Delta H_\text{Fusion}$ .

$E(\text{Vaporization}) = n \cdot \Delta H_\text{Vaporization}$ .

where $n$ is the number of moles of the substance.

Look up the relative atomic mass of oxygen and hydrogen from a modern periodic table:

H: 1.008,
O: 15.999.

Hence the molar mass of water:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g \cdot mol^{-1}$ .

Number of moles of $\rm H_2O$ molecules in $\rm 13.0\; g$ :

$\displaystyle n = \frac{m}{M} \approx 0.721621\; \rm mol$ .

$\begin{aligned}& &&\text{Energy required for phase changes} \cr &=&& n \cdot \Delta H_\text{Fusion} \cr & &&+n \cdot \Delta H_\text{Vaporization} \cr & = &&0.721621 \times 6.02 + 0.721621 \times 40.7 \cr & = &&33.7\; \rm kJ \end{aligned}$

<h3>Energy required for all five steps, combined</h3>

$5.969\; \rm kJ + 33.7\; \rm kJ \approx 39.7\; \rm kJ$ .

8 0

3 years ago

Scientists believe that the Earth's orbit gradually changes from being elliptical to being nearly circular and then back to elli

IRISSAK [1]

Answer:

solar energy received at the Earth's closest and farthest locations from the sun

Explanation:

The largest effect on climate is due to changes in received solar radiation. Changes in gravity or magnetic field have not been shown to have any significant effect on climate. Day/night differences are due to the tilt of the earth's axis, and have nothing to do with earth's orbit eccentricity.

The more circular the Earth's orbit, the less the difference in solar energy received at the Earth's closest and farthest locations from the sun.

3 0

3 years ago

Na2S2O3 + AgBr NaBr + Na3ſAg(S203)2] What is

gtnhenbr [62]

Answer:

Mass of NaBr produced = 23.67 g

Explanation:

Given data:

Mass of AgBr = 42.7 g

Mass of NaBr produced = ?

Solution:

Chemical equation:

2Na₂S₂O₃ + AgBr → NaBr + Na₃(Ag(S₂O₃)₂

Number of moles of AgBr:

Number of moles = mass/molar mass

Number of moles = 42.7 g/ 187.7 g/mol

Number of moles = 0.23 mol

now we will compare the moles of AgBr with NaBr.

AgBr : NaBr

1 : 1

0.23 : 0.23

Mass of NaBr:

Mass = number of moles × molar mass

Mass = 0.23 mol × 102.89 g/mol

Mass = 23.67 g

8 0

3 years ago