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Ludmilka [50]
3 years ago
10

The specific heat of a substance is 0.215 J/g°C. How much energy is required to raise the temperature of 20 g of the substance f

rom 72°C to 88°C?
Physics
1 answer:
sergey [27]3 years ago
6 0

Taking into account the definition of calorimetry and sensible heat, the amount of energy required is 68.8 J.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

In this way, between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body.

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • c is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Energy required in this case</h3>

In this case, you know:

  • Q= ?
  • c= 0.215 \frac{J}{gC}
  • m= 20 g
  • ΔT= Tfinal - Tinitial= 88 C - 72 C= 16 C

Replacing in the definition of sensible heat:

Q = 0.215 \frac{J}{gC}× 20 g× 16 C

Solving:

<u><em>Q=68.8 J</em></u>

Finally, the amount of energy required is 68.8 J.

Learn more about calorimetry:

<u>brainly.com/question/11586486?referrer=searchResults</u>

<u>brainly.com/question/24724338?referrer=searchResults</u>

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A 28 cm hammer is used to pull a nail. If 28.5N of force is
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7.23 Nm

Explanation:

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3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
4 years ago
A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change
lord [1]

The mass that must be added is 0.628 kg

Explanation:

The period of a mass-spring system is given by

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

For the initial mass-spring system in the problem, we have

m = 0.500 kg

T = 1.36 s

Solving for k, we find the spring constant:

k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m

In the second part, we want the period of the same system to be

T = 2.04 s

Therefore, the mass on the spring in this case must be

m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg

Therefore, the mass that must be added is

\Delta m = 1.128 - 0.500 = 0.628 kg

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Answer:

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Explanation:

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?

the magnetic field Bnet=\sqrt{b1^2+b2^2}

the magnetic field due this long wire is given by

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B2=∨I2/(2\pi *R2)............................2

Bnet=\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}.......................3

Bnet=v/2*pi\sqrt{(I1/R1)^2+(i2/R2)^2}

Bnet=4*pi*10^-7/(2\pi)\sqrt{(43/1.7)^2+(41/29.5)^2}

Bnet=0.0000002*(641.72)^.5

Bnet=1.006*10^-6T

8 0
4 years ago
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