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konstantin123 [22]
3 years ago
9

Determine the specific heat of a certain metal if a 450 gram sample of it loses 34 500 Joules of heat as its temperature starts

at 150 oC and drops to 53 oC.
Physics
1 answer:
slamgirl [31]3 years ago
6 0

Answer:

c = 0.4356 J/gK

Explanation:

Given the following data;

Mass = 450 grams

Initial temperature, T1 = 150°C

Final temperature, T2 = 53°C

Quantity of heat = 34500 Joules

To find the specific heat capacity of the metal;

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 53 - 150

dt = -97°C

Converting the temperature in Celsius to Kelvin, we have;

dt = 273 + (-97) = 176 Kelvin

Making c the subject of formula, we have;

c = \frac {Q}{mdt}

Substituting into the equation, we have;

c = \frac {34500}{450*176}

c = \frac {34500}{79200}

c = 0.4356 J/gK

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In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

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