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4 years ago

8

A sample of solid iron is heated with an electrical coil. If 85.7 Joules of energy are added to a 14.3 gram sample initially at

24.7 °C, what is the final temperature of the iron?

1 answer:

Masja [62]4 years ago

7 0

Answer:

The final temperature of the iron is:- 38.0 °C

Explanation:

The initial temperature = 24.7 °C

Let Final temperature = T °C

Using,

Q = m C ×ΔT

Where,

Q is the heat absorbed by the iron = 85.7 J

C gas is the specific heat of the iron = 0.45 J/g °C

m is the mass of iron = 14.3 g

ΔT is the change in temperature. ( T - 24.7 ) °C

Applying the values as:

<u>Q = m C ×ΔT </u>

85.7 J = 14.3 g × 0.45 J/g °C × ( T - 24.7 ) °C

Solving for T, we get that:-

T = 38.0 °C

<u>The final temperature of the iron is:- 38.0 °C</u>

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Answer:

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Explanation:

Using the derived expression from Arrhenius Equation

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Given that:

time $t_1$ = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time $t_2$ = 10.6 hours

Temperature $T_1$ = 0° C = (0+273 )K = 273 K

Temperature $T_2$ = 30° C = (30+ 273) = 303 K

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Since $(\frac{k_2}{k_1}=\frac{t_2}{t_1})$

Then we can rewrite the above expression as:

$In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

$In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})$

$2.934 = \frac{E_a}{8.314}(\frac{30}{82719})$

$2.934 = \frac{30E_a}{687725.766}$

$30E_a = 2.934 *687725.766$

$E_a = \frac{2.934 *687725.766}{30}$

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3 years ago

Gold has an atomic weight of 196.97 g/mol, and a density of 19.3 g/cm3. Approximately how many atoms are in a spherical gold nan

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<u>Answer:</u> The number of atoms in spherical gold nano particle are $26.918\times 10^4$

<u>Explanation:</u>

To calculate the volume of sphere, we use the equation:

$V=\frac{4}{3}\pi r^3$

where,

r = radius of the sphere = $10nm=10^{-6}cm$ (Conversion factor: $1nm=10^{-7}cm$ )

Putting values in above equation, we get:

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To calculate mass of of the substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of gold = $4.19\times 10^{-18}cm^3$

Density of gold = $19.3g/cm^3$

Putting values in above equation, we get:

$19.3g/cm^3=\frac{\text{Mass of gold}}{4.19\times 10^{-18}}\\\\\text{Mass of gold}=8.0867\times 10^{-17}g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of gold = $8.8067\times 10^{-17}g$

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Putting values in above equation, we get:

$\text{Moles of gold}=\frac{8.8067\times 10^{-17}g}{196.97g/mol}=4.47\times 10^{-9}moles$

According to mole concept:

1 mole of an element contains $6.022\time 10^{23}$ number of atom

So, $4.47\times 10^{-19}$ moles of gold will contain = $4.47\times 10^{-19}\times 6.022\times 10^{23}=26.918\times 10^4$ number of atoms.

Hence, the number of atoms in spherical gold nano particle are $26.918\times 10^4$

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