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1 year ago

How much heat is needed to melt 100.0 grams of ice that is already at 0°C?

1 answer:

4 0

Taking into account the definition of calorimetry and latent heat, the heat needed to melt 100.0 grams of ice that is already at 0°C is 33,400 J (option A).

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

In this case, the heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

<h3>Heat needed to melt ice</h3>

In this case, you have to melt the ice into liquid water.Being the specific heat of melting of ice is 334 J/g, the heat needed to melt 100 grams of ice is calculated as:

Q= 100 grams× 334 J/g

Solving:

In summary, the heat needed to melt 100.0 grams of ice that is already at 0°C is 33,400 J (option A).

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6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)

scoray [572]

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

<h3>Mass of water required</h3>

In this case, you know:

Heat= 92.048 kJ
Mass of water = ?
Initial temperature of water= 34 ºC
Final temperature of water= 100 ºC
Specific heat of water = 4.186 $\frac{J}{gC}$

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 $\frac{J}{gC}$ × m× (100 °C -34 °C)

92.048 kJ = 4.186 $\frac{J}{gC}$ × m× 66 °C

m= 92.048 kJ ÷ (4.186 $\frac{J}{gC}$ × 66 °C)

<u><em>m= 0.333 grams</em></u>

<h3>Moles of water required</h3>

Being the molar mass of water 18 $\frac{g}{mole}$ , that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:

$amount of moles=0.333 gramsx\frac{1 mole}{18 grams}$

<u><em>amount of moles= 0.0185 moles</em></u>

Finally, 0.0185 moles of water are required.

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8 0

3 years ago

The following reaction can be written as the sum of two reactions, one of which relates to ionization energy and one of which re

Karo-lina-s [1.5K]

A)
First ionization energy of Li:
Li(g) + 520 kJ/mol → Li⁺(g) + e⁻

B)
Electron affinity of Fluorine:
F(g) + e⁻ → F⁻(g) + 328 kJ/mol

8 0

3 years ago

Which best describes the reducing agent in the reaction below? Cl2(aq) + 2Br(aq) 2Cl(aq) + Br2(aq) Bromine (Br) loses an electro

SIZIF [17.4K]

Answer:

Bromine (Br) loses an electron, so it is the reducing agent.

Explanation:

A reducing agent also called a reducer, is known to be an electron donor. A reducing agent is oxidized, because it loses electrons in the redox reaction.

A oxidising agent also called a oxidant or oxidiser, is known to be an electron acceptor. A oxidising agent is reduced, because it gains electrons in the redox reaction.

Cl2(aq) + 2Br-(aq) --> 2Cl-(aq) + Br2(aq)

Half ionic equations,

Cl2(aq) + 2e- --> 2Cl-(aq)

2Br-(aq) --> Br2(aq) + 2e-

Reducing agent = Br-

Oxidizing agent = Cl2

4 0

3 years ago

Read 2 more answers

Who was the scientist that coined the term cell?

Doss [256]

Robert Hooke was the scientist that coined the term cell

5 0

3 years ago

what is the amount of heat energy released when 50.0 grams of water is cooled from 20.0 degrees Celcius to 10.0 degrees celcius

andre [41]

Answer:

The amount of heat released when 50 g of water cooled from 20°C to 10°C will be equal to - 2093 J.

Explanation:

Given data:

Mass of water = 50 g

Initial temperature= T1 = 20°C

Final temperature= T2 = 10°C

Specific heat of water= c = 4.186 J/g. °C

Amount of heat released = Q= ?

Solution:

Formula:

Q = m. C. ΔT

ΔT = T2 - T1

ΔT = 10°C - 20°C

ΔT = -10°C

Now we will put the values in formula.

Q = m. C. ΔT

Q = 50 g . 4.186 J/g. °C . -10°C

Q = - 2093 J

The amount of heat released when 50 g of water cooled from 20°C to 10°C will be equal to - 2093 J.

6 0

3 years ago

Read 2 more answers