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alexandr402 [8]
3 years ago
8

A sample of an unknown substance has a mass of 0. 158 kg. If 2,510. 0 J of heat is required to heat the substance from 32. 0°C

to 61. 0°C, what is the specific heat of the substance? Use q equals m C subscript p Delta T. 0. 171 J/(gi°C) 0. 548 J/(gi°C) 15. 9 J/(gi°C) 86. 6 J/(gi°C).
Physics
1 answer:
guapka [62]3 years ago
4 0

The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

The specific heat capacity of a substance can be calculated using the following formula:

Q = m × c × ∆T

Where;

  • Q = quantity of heat absorbed (J)
  • c = specific heat capacity (4.18 J/g°C)
  • m = mass of substance
  • ∆T = change in temperature (°C)

According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:

2510 = 158 × c × (61°C - 32°C)

2510 = 4582c

c = 2510 ÷ 4582

c = 0.5478 J/g°C

Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.

Learn more about specific heat capacity at: brainly.com/question/2530523

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A plane is traveling with an air velocity of 720 kilometers/hour due east. It experiences a headwind of 16 kilometers/hour. Find
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What are non-clastic?
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4 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

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Answer:

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Explanation:

so to be exact galixes do make up most of the volume of space because galaxies are what make up space!

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