All categories

2 years ago

How much heat is required to raise the temperature of 67.0 g of water from its melting point to its boiling point?

Chemistry

1 answer:

vodka [1.7K]2 years ago

3 0

When 67 g of water is heated from its melting point to its boiling point, it takes 28006 J of heat.

<h2>Relationship between heat production and temperature change</h2>

A way to numerically relate the quantity of thermal energy acquired (or lost) by a sample of any substance to that sample's mass and the temperature change that results from that is provided by specific heat capacity.

The following formula is frequently used to describe the connection between these four values.

q = msΔT

where, q = the amount of heat emitted or absorbed by the thing

m = the object's mass = 67 gm

s = a specific heat capacity of the substance = 4.18 J/gC

ΔT = the resultant change in the object's temperature = 373.15 -273.15K= 100 k

q = 67 * 4.18 * 100 J

⇒q = 28006 J

Therefore it is concluded that 67 g of water takes 28006 J of heat from its melting point to reach its boiling point.

Learn more about thermal energy here:

brainly.com/question/3022807

#SPJ1

You might be interested in

How many molecules are in 3 moles of potassium bromide (KBr)

sattari [20]

Answer:

Your strategy here will be to use the molar mass of potassium bromide,

KBr

, as a conversion factor to help you find the mass of three moles of this compound.

So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations,

, and bromide anions,

−

. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.

Use the periodic table to find the molar masses of these two elements. You will find

For K:

39.0963 g mol

−

For Br:

79.904 g mol

−

To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements

M KBr

39.0963 g mol

−

79.904 g mol

−

≈

119 g mol

−

So, if one mole of potassium bromide has a mas of

119 g

m it follows that three moles will have a mass of

moles KBr

⋅

molar mass of KBr



119 g

mole KBr

357 g

You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

mass of 3 moles of KBr

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

360 g

∣

−−−−−−−−−

Explanation:

answer: 360 g 

6 0

3 years ago

Do all cells have the same internal structures

professor190 [17]

Answer:

Explanation:

Each eukaryotic cell has a plasma membrane, cytoplasm, a nucleus, ribosomes, mitochondria, peroxisomes, and in some, vacuoles; however, there are some striking differences between animal and plant cells. ... Animal cells each have a centrosome and lysosomes, whereas plant cells do not.

8 0

3 years ago

Please answer asap MUCH NEEDED!!

gulaghasi [49]

Answer:

I think its C I am sorry if I am wrong

3 0

3 years ago

Energy leaves the sun as light from region called the?

tino4ka555 [31]

Late answer:
Chromosphere.

4 0

3 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

3 years ago

Read 2 more answers