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Minchanka [31]
3 years ago
15

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

Physics
1 answer:
sweet-ann [11.9K]3 years ago
5 0

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

A)

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

Q=mC\Delta T

where

m is the mass of the substance

C is the specific heat capacity

\Delta T is the change in temperature

In this problem:

m = 1 g is the mass of water

C=1 cal/g^{\circ}C is  the specific heat capacity of water

\Delta T=1^{\circ}C is the change in temperature

So, the heat needed is

Q=(1)(1)(1)=1 cal

B)

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

Q=m\lambda_f

where

m is the mass of the substance

\lambda_f is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: m=1g

- Specific latent heat of fusion of ice: \lambda_f=80 cal/g

So, the heat needed is

Q=(1)(80)=80 cal

C)

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

Q=m\lambda_v

where

m is the mass of the substance

\lambda_v is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: m=1g

- Specific latent heat of vaporization of water: \lambda_v=540 cal/g

So, the heat needed is

Q=(1)(540)=540 cal

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Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

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When the object is suspended, the gravitational force is balanced by the spring force as :

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k=\dfrac{3.74\times 9.8}{0.0161}

k = 2276.52 N/m

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f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2f^2}

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Answer:

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Explanation:

First we must understand the data given in the problem:

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F = force = 20000[N]

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From newton's second we know that the sum of forces must be equal to the product of mass by acceleration.

F = m*a\\20000 = 800*a\\a = 20000/800\\a = 25 [m/s^2]

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The key to using this equation is to clarify that the initial velocity is zero since the body is at rest, otherwise the initial velocity would be an initial data.

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Another way of solving this problem is by means of the definition of work and kinetic energy, where work is defined as the product of the force by the distance.

W =F*d

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W = 200000[J]

Kinetic energy is equal to work, therefore the value calculated above is equal to:

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