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lisabon 2012 [21]
3 years ago
15

How much heat transfer (in kcal) is required to raise the temperature of a 0.850 kg aluminum pot containing 2.00 kg of water fro

m 45.0°C to the boiling point and then boil away 0.700 kg of water?
Physics
1 answer:
hram777 [196]3 years ago
8 0

To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T  +m_w L_v

Here,

m_{Al}= Mass of Aluminum

C_p= Specific Heat of Aluminum

C_w= Specific Heat of Water

m_w = Mass of water

L_v =Latent of Vaporization

Replacing,

\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)

Converting,

\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})

\Delta Q = 440.409kCal

Therefore is required 440.409kCal

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Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a
marishachu [46]

Answer:

-9.46 N

Explanation:

In order to get the value of the x component of the resultant force, we need to get the value of the x component of each force.

This value will be the projection of the force vector, on the x-axis.

For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₁ₓ, and r = F₁

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º-55º = 125º

⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N

We can repeat the process for F₂, as follows:

For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₂ₓ, and r = F₂

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º + 53.3º = 233.3º

⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N

The total component of both forces along the x axis, can be found just adding both components, as follows:

Fₓ = F₁ₓ + F₂ₓ = -5.28 N + -4.18 N = -9.46 N

5 0
2 years ago
A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of
Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

\eta = 1-\frac{T_L}{T_H}

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

\eta = 1-\frac{T_L}{T_H}

\eta = 1-\frac{293}{713}

\eta = 0.589

Therefore the maximum possible efficiency the car can have is 58.9%

4 0
2 years ago
Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as se
alisha [4.7K]

Answer with Explanation:

Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.

For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as  pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference

While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.  

3 0
3 years ago
How is volume and temperature related?
ololo11 [35]

Charles' Law: The Temperature-Volume Law. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. As the volume goes up, the temperature also goes up, and vice-versa.

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The true brightness of an object is called its
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8 0
3 years ago
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