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3 years ago

What is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C

Chemistry

2 answers:

SOVA2 [1]3 years ago

5 0

Answer:

4702.5 J/g*k

Explanation:

Depending on if it is water as a solid liquid or gas. I used water as a liquid when I solved it. J=(75g)(4.18 J/g*k)(15K)

Alenkasestr [34]3 years ago

4 0

Answer:

4.704625 kJ (or 4704.625 J) is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C.

Explanation:

The measurement and calculation of the amounts of heat exchanged by a body or system is called calorimetry.

In this way, there is a direct proportionality relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as its mass, and the product of specific heat is produced by the body's mass. So, the equation that allows to calculate heat exchanges is:

Q = c * m * T

Where Q is the heat exchanged for a body of mass m, constituted by a specific heat substance c and where T is the temperature variation.

In this case:

The specific heat capacity (c) of water is 4,181 J / g°C
m=75 g
T=35°C - 20°C ⇒ T= 15°C

Then:

$Q= 4.181 \frac{J}{g C} * 75 g *15 C$

Q≅4704.625 J

Being 1 kJ = 1000 J:

Q=4.704625 kJ

Finally, 4.704625 kJ (or 4704.625 J) is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C

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oxidation-reduction reaction or redox reaction.

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2 years ago

Calculate the amount of heat needed to raise 190.0 g of water vapor at 118 C to 150. C

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<h3>Answer:</h3>

12,281.6 Joules

<h3>Explanation:</h3>

The amount of heat energy is calculated by multiplying the mass of a substance by its specific heat capacity and then by the change in temperature.

That is, Q = m × c ×Δt

In our case;

Mass of water vapor, m = 190.0 g

Change in temperature, Δt = 150°C - 118°C = 32°C

Taking the specific heat capacity of water vapor = 2.02 J/g°C

Therefore;

Amount of heat = 190.0 g × 2.02 J/g°C × 32°C

= 12,281.6 Joules

Hence, the amount of heat needed is 12,281.6 Joules

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2 years ago

Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit

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Answer:

(a): The theoretical yield of silicon is 72.33 kg.

(b): The percent yield of the reaction is 91.25 %.

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $SiO_2$ :

Given mass = 155.0 kg = 155000 g (Conversion factor: 1 kg = 1000 g)

Molar mass = 60 g/mol

Putting values in equation 1:

$\text{Moles of }SiO_2=\frac{155000g}{60g/mol}=2583.3mol$

For carbon:

Given mass = 78.2 kg = 78200 g

Molar mass = 12 g/mol

Putting values in equation 1:

$\text{Moles of carbon}=\frac{78200g}{12g/mol}=6516.67mol$

The chemical equation for the reaction of silicon dioxide and carbon follows:

$SiO_2+2C\rightarrow Si+2CO$

By stoichiometry of the reaction:

1 mole of $SiO_2$ reacts with 2 moles of carbon

So, 2583.3 moles of $SiO_2$ will react with = $\frac{2}{1}\times 2583.3=5166.4mol$ of carbon

As the given amount of carbon is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, $SiO_2$ is considered a limiting reagent because it limits the formation of the product.

For (a):

By stoichiometry of the reaction:

1 mole of $SiO_2$ produces 1 mole of silicon

So, 2583.3 moles of $SiO_2$ will produce = $\frac{1}{1}\times 2583.3=2583.3mol$ of silicon

Since the molar mass of silicon = 28 g/mol

Putting values in equation 1:

$\text{Mass of Si}=2583.3mol\times 28g/mol=72332.4g=72.33 kg$

Hence, the theoretical yield of silicon is 72.33 kg.

For (b):

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Measured value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Measured value of silicon = 66.0 kg

Theoretical value of silicon = 72.33 kg

Putting values in equation 1:

$\% \text{yield}=\frac{66.0kg}{72.33kg}\times 100\\\\\% \text{yield}=91.25 \%$

Hence, the percent yield of the reaction is 91.25 %.

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2 years ago

An ideal diatomic gas starting at room temperature T1 = 300 K and atmospheric pressure p1 = 1.0 atm is compressed adiabatically

tangare [24]

Answer:

The final temperature of the given ideal diatomic gas: T₂ = 753.6 K

Explanation:

Given: Atmospheric pressure: P = 1.0 atm

Initial Volume: V₁ , Final Volume: V₂ = V₁ (1/10)

⇒ V₁ / V₂ = 10

Initial Temperature: T₁ = 300 K, Final temperature: T₂ = ? K

For a diatomic ideal gas: γ = 7/5

For an adiabatic process:

$V^{\gamma-1 }T = constant$

$V_{1}^{\gamma-1 }T_{1} = V_{2}^{\gamma-1 }T_{2}$

$\left [\frac{V_{1}}{V_{2}} \right ]^{\gamma-1 } = \frac{T_{2}}{T_{1}}$

$\left [10 \right ]^{\frac{7}{5}-1 } = \frac{T_{2}}{300 K}$

$\left [10 \right ]^{\frac{2}{5} } = \frac{T_{2}}{300 K}$

$2.512 = \frac{T_{2}}{300 K}$

$T_{2} = 753.6 K$

Therefore, the final temperature of the given ideal diatomic gas: T₂ = 753.6 K

8 0

2 years ago