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3 years ago

How much energy does it take to raise the temperature of 80g of aluminium 15 degree Celsius

2 answers:

6 0

Answer :The specific heat of aluminium tells you the amount of energy needed to increase the temperature of 1 g of aluminium by 1∘C . You can thus say that in order to increase the temperature of 1 g of aluminium by 1∘C , you need to supply it with 0.214 cal of heat.

Explanation:

irinina [24]3 years ago

5 0

Answer:

$\boxed {\boxed {\sf 1080 \ Joules}}$

Explanation:

We are asked to calculate the energy needed to raise the temperature of 80 grams of aluminum by 15 degrees Celsius.

We are given the mass and change in temperature, so we will use the following formula:

$q=mc\Delta T$

The mass is 80 grams, the change in temperature is 15 degrees Celsius, and aluminum's specific heat capacity is 0.9 Joules per gram degree Celsius.

m=80 g
ΔT= 15°C
c= 0.9 J/g °C

Substitute the values into the formula.

$q= (80 \ g)(0.9 \ J/g \textdegree C)(15 \textdegree C)$

Multiply the first two numbers together. The units of grams will cancel out.

$q= (80 \ g * 0.9 \ J/g \textdegree C)(15 \textdegree C)$

$q= (80 * 0.9 \ J/\textdegree C)(15 \textdegree C)$

$q= 72 \ J / \textdegree C (15 \textdegree C)$

Multiply again. This time, the units of degrees Celsius cancel.

$q= 72 \ J * 15$

$q= 1080 \ J$

Raising the temperature of 80 grams of aluminum by 15 degrees Celsius requires <u>1080 Joules of energy.</u>

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What is the difference between thermal energy and temperature?

igor_vitrenko [27]

The hotter an object is, the faster the motion of the molecules inside it. Thus, the heat of an object is the total energy of all the molecular motion inside that object. Temperature, on the other hand, is a measure of the average heat or thermal energy of the molecules in a substance.

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3 years ago

An unknown compound containing only C and H was burnt, yielding 10.2 g of CO2 and 6.3 g of H2O. With a molecular weight of about

Lisa [10]

Answer:

$C_2H_6$

Explanation:

Hello.

In this case, we can see that the mass of carbon of the unknown compound comes from the yielded mass of carbon dioxide, thus, we compute the moles of carbon as follows:

$m_C=10.2gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1mol C}{1molCO_2}=0.232 molC$

Moreover, the mass of hydrogen comes from the yielded water, therefore we can also compute the moles of water:

$m_H=6.3gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH_2}{1molH_2O} =0.7molH$

Then, to find the subscripts in the empirical formula, we divide by the moles of carbon as the smallest:

$C:\frac{0.232}{0.232}=1\\ \\H:\frac{0.7}{0.232}=3$

Whose molar mass is:

$M_{CH_3}=12+1*3=15g/mol$

Thus, the ratio of the molecular formula to the empirical formula is:

$\frac{30}{15}=2$

Therefore, the molecular formula is twice the empirical formula:

$C_2H_6$

Which is actually ethane.

Regards.

3 0

3 years ago

1. Lithium, water, edible salt, chalk, Carbon, Lime, Nitrogen, Potassium, Oxygen,

Kaylis [27]

Answer:

Explanation:

An atom is the smallest unit of an element that can take part in a chemical reaction. Atoms (and there corresponding symbols) mentioned in the question are

Lithium ⇒ Li

Carbon ⇒ C

Nitrogen ⇒ N

Potassium ⇒ K

Oxygen ⇒ O

Iron ⇒ Fe

Chlorine ⇒ Cl

A compound is substance that contains two or more atoms that are chemically combined and can be represented with a chemical formula. The compounds (and there corresponding formula) mentioned in the question are

Water ⇒ H₂O

Edible salt (sodium chloride) ⇒ NaCl

Chalk (calcium carbonate) ⇒ CaCO₃

Lime (calcium oxide) ⇒ CaO

Iodides (such as sodium iodide and potassium iodide) ⇒ NaI and KI respectively

8 0

3 years ago

You are preparing some solutions for your instructor and have been asked to prepare 3.00 L of a 0.250M sodium hydroxide solution

grigory [225]

Answer:

C. 30.0 g NaOH and add water until the final solution has a volume of 3.00 L.

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

from the question ,

M = 0.250M

V = 3.00 L

M = n / V

n = M * v

n = 0.250M * 3.00 L = 0.75 mol

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

n = 0.75 mol NaOH

m = molecular mass of NaOH = 40 g/mol

n = w / m

w = n * m

w = 0.75 mol * 40 g/mol = 30.0 g

Hence , by using 30.0 g of NaOH and dissolving it to make up the volume to 3 L , a solution of 0.250 M can be prepared .

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3 years ago

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Allushta [10]

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