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3 years ago

Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.

1 answer:

7 0

Given:
Iron, 125 grams T
1 = 23.5 degrees Celsius, T2 = 78 degrees Celsius.

Required:
Heat produced in kilojoules

Solution:
The molar mass of iron is 55.8 grams per mole. SO we need to change the given mass of iron into moles.

Number of moles of iron = 125 g/(55.8 g/mol) = 2.24 moles

Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2.24 moles (8.314 Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
Q (heat) = 1014.97 Joules or 1.015 kilojoules This is the amount of heat produced in warming 125 g f iron.

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A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of

inna [77]

The mass of the copper cylinder is 177065.856g

Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 $\frac{g}{c m^{3}}$

To find:

Mass of the copper cylinder

Step by Step by explanation:

Solution:

According to the formula, Mass can be calculated as

$\rho=\frac{m}{v}$ and from this

$m=\rho \times v$

Where, m=mass of the cylinder

$\rho$ =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

$v=\pi r^{2} h$

Where $\pi$ =3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, $v=3.14 \times 12^{2} \times 44$

v=3.14 \times 144 \times 44

$v=19895.04 \mathrm{cm}^{3}$

And we know that, $m=\rho \times v$

Substitute the known values in the above equation we get

$m=8.90 \times 19895.04$

m=177065.856g or 177.065kg

Result:

Thus the mass of the copper cylinder is 177065.856g

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