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nadya68 [22]
2 years ago
11

A sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 1

00.0°c, what is the specific heat of the substance?
Physics
1 answer:
hammer [34]2 years ago
5 0

The specific heat of the substance will be 0.129 J/g°C.

<h3>What is specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

The given data in the problem is;

Q is the amount of energy necessary to raise the temperature = 3,000.0 j

M is the mass=  0.465 kg.

Δt is the time it takes to raise the temperature.=50°c

s stands for specific heat capacity=?

Mathematically specific heat capacity is given by;

\rm Q= MC \triangle t \\\\ C = \frac{Q}{M\triangle t} \\\\ C = \frac{3000}{0.465 \triangle 50} \\\\ C =129.0 J/Kg^0C \\\\ C= 0.129 J/g^0C

Hence the specific heat of the substance will be 0.129 J/g°C.

To learn more about the specific heat capacity refer to the link brainly.com/question/2530523

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
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A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

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