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s344n2d4d5 [400]
4 years ago
8

Which metal would cause the greatest increase in the temperature of the water in the calorimeter: the one with the highest speci

fic heat, or the one with the lower specific heat?
Physics
1 answer:
masha68 [24]4 years ago
4 0
<h3><u>Answer;</u></h3>

the one with the highest specific heat

<h3><u>Explanation;</u></h3>
  • The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.
  • The specific heat of water is 1 calorie/gram °C or 4.186 joule/gram °C which is higher than any other common substance.
  • <em><u>A metal with the highest specific heat will cause the greatest increase in temperature of water in a calorimeter because the metal would hold more heat, and then transfer the greater quantity of heat to the water.</u></em>
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scZoUnD [109]

Answer:

28 grams

Explanation:

The equation for the reaction is

3H(2) + N(2) -> 2NH(3)

Then we have.

The molar mass, M of ammonia is 17 g/mol.

34 grams of ammonia, NH3 then would be

34 g / 17 g/mol

= 2 moles

2 moles of ammonia will be obtained from

(2 * 1) / 2

​= 1 mole of nitrogen

The molar masses of nitrogen is 28 g/mol

2 moles of nitrogen corresponds to 1 * 28 = 28 grams.

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Explanation:

8 0
3 years ago
Read 2 more answers
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm

the radius of the trajectory of the electron is 10.1 cm

6 0
4 years ago
Two strong magnets were brought close to each other. They were repelling each other. Explain what must have happened.​
r-ruslan [8.4K]

When two magnets are brought together, the opposite poles will attract one another, but the like poles will repel one another. This is similar to electric charges. Like charges repel, and unlike charges attract.

pls. mark brainliest am. dyning for it

8 0
3 years ago
Read 2 more answers
A 12-V battery is connected to an air-filled capacitor that consists of two parallel plates,
zloy xaker [14]

Answer:

E = 4000 V / m

U = 1.92*10^-18 J

C' = 4.71 pF

1.2 times greater with di-electric

Explanation:

Given:-

- The potential difference between plates, V = 12 V

- The area of each plate, A = 7.6 cm^2

- The separation between plates, d = 0.3 cm

- The charge of the proton. q = 1.6*10^-19 C

- The initial velocity of proton, vi = 0 m/s

Solution:-

- The electric field ( E ) between the parallel plates of the air-filled capacitor is determined from the applied potential difference by the battery on the two ends of the plates.

- The separation ( d ) between the two plates allows the charge to be stored and the Electric field between two charged plates would be:

                          E = V / d

                          E = 12 / 0.003

                          E = 4,000 V/m ... Answer

- The amount of electrostatic potential energy stored between the two plates is ( U ) defined by:

                         U = q*E*d

                         U = (1.6 x10^-19)*(4000)*(0.003)

                         U = 1.92*10^-18 J  ... Answer

- The electrostatic energy stored between plates is ( U ) when the proton moves from the positively charges plate to negative charged plate the energy is stored within the proton.

- A slab of di-electric material ( Teflon ) is placed between the two plates with thickness equal to the separation ( d ) and Area similar to the area of the plate ( A ).

- The capacitance of the charged plates would be ( C ):

                        C = k*ε*A / d

Where,

            k: the di-electric constant of material = 2.1

            ε: permittivity of free space = 8.85 × 10^-12

- The new capacitance ( C' ) is:

                      C' = 2.1*(8.85 × 10^-12) *( 7.6 / 100^2 ) / 0.003

                      C' = 4.71 pF

- The new total energy stored in the capacitor is defined as follows:

                     U' = 0.5*C'*V^2

                     U' = 0.5*(4.71*10^-12)*(12)^2

                     U' = 3.391 * 10^-10 J

- The increase in potential energy stored is by the amount of increase in capacitance due to di-electric material ( Teflon ). The di-electric constant "k" causes an increase in the potential energy stored before and after the insertion.

- Hence, the new potential energy ( U' ) is " k = 2.1 " times the potential energy stored in a capacitor without the di-electric.

                     

4 0
3 years ago
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