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Alexxx [7]
2 years ago
6

A molten sample of 1.00kg of iron with a specific heat of 0.385J/g.K at 1000.K is immersed in a sample of water. If the water ab

sorbs 270 kJ of heat what is the final temperature of the iron?
Chemistry
1 answer:
Schach [20]2 years ago
6 0

Answer:

Final T = 298.7K

Explanation:

In this problem, the heat given for the iron is equal to the heat that water absorbs. Using the equation:

Q = C*m*ΔT

<em>Where Q is heat given = 270000J</em>

<em>C is specific heat of substance = 0.385J/gK</em>

<em>m is mass of iron = 1000g</em>

<em>ΔT is change in temperature = Initial temperature - Final temperature.</em>

<em />

270000J = 0.385J/gK*1000g*(1000K-Final temperature)

701.3J/K = 1000K - Final T

<h3>Final T = 298.7K</h3>
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An object's velocity can be described by its acceleration and direction.<br><br> True<br> False
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3 years ago
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Explanation:

6. Where on the graph does adding heat energy NOT raise the temperature?

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4 0
3 years ago
Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold
Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}

where,

N_{Au} = number of gold atoms per cubic centimeters

N_A = Avogadro's number = 6.022\times 10^{23}atoms/mol

C_{Au} = Mass percent of gold in the alloy = 42 %

\rho_{Au} = Density of pure gold = 19.32g/cm^3

\rho_{Ag} = Density of pure silver = 10.49g/cm^3

M_{Au = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3

Hence, the number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

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