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3 years ago

How much heat is absorbed by a 47g iron skillet when its temperature rises from 12oC to 20oC?

Physics

1 answer:

jolli1 [7]3 years ago

6 0

Answer

169.2 J

Explanation

Given in the question,

mass of iron = 47g

specific heat capacity of iron = 0.450 (J/g 0C)

initial temperature = 12° C

final temperature = 20° C

The energy q needed to increase an object of mass m and specific heat capacity c by a temperature θ is given by:

q = mcΔt

q = 47(0.45)(20-12)

q = 169.2 J

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Answer:

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Explanation:

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3 years ago

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3 years ago

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Minchanka [31]

Answer:

r = 0.11 m

Explanation:

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Where:

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v: is the proton's velocity

r: is the radius of the proton's orbit

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Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$ (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$ (3)

Where:

K: is kinetic energy

U: is electrostatic potential energy

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Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

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3 years ago

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2 years ago