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alexira [117]
3 years ago
5

Calculate the amount of heat needed to raise 1.0 kg of ice at -20 degrees Celsius to steam at 120 degree Celsius

Physics
1 answer:
CaHeK987 [17]3 years ago
4 0

Answer:

801.1 kJ

Explanation:

The ice increases in temperature from -20 °C to 0 °C and then melts at 0 °C.

The heat required to raise the ice to 0 °C is Q₁ = mc₁Δθ₁ where m =  mass of ice = 1 kg, c₁ = specific heat capacity of ice = 2108 J/kg°C and Δθ₁ = temperature change. Q₁ = 1 kg × 2108 J/kg°C × (0 - (-20))°C = 2108 J/kg°C × 20  °C = 4216 J

The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J

The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m =  mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100  °C = 418700 J

The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J

The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m =  mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20  °C = 39920 J

The total amount of heat Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 4216 J + 336000 J

+ 418700 J + 2260 J + 39920 J = 801096 J ≅ 801.1 kJ

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3 years ago
A 55 kg track and field athlete has an average power output of 5.4 kW during the 200 meter dash. How quickly did she finish the
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The time taken for the athlete to finish the race is 20 s (Option A)

<h3>What is power? </h3>

Power is simply defined as the rate at which work is done. It can be expressed mathematically as

Power (P) = work (W) / time (t)

But

Work = weight × distance

Therefore,

Power = (weight × distance ) / time

<h3>How to determine the time </h3>
  • Mass (m) = 55 Kg
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Weight = mg = 55 × 9.8 = 539 N
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Power = (weight × distance ) / time

5400 = (539 × 200) / t

5400 = 107800 / t

Cross multiply

5400 × t = 107800

Divide both side by 5400

t = 107800 / 5400

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5 0
1 year ago
A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the
ser-zykov [4K]

Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

           y is the difference in weight which is   =1.10-0.50=0.6m

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

                       = 8.17 N/m

The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

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