Conservation of Energy and Temperature Change A crate of fruit with a mass of 39

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Conservation of Energy and Temperature Change

A crate of fruit with a mass of 39.5 kg and a specific heat of 3800 J/(kgK) slides 7.00 mdown a ramp inclined at an angle of 40.0 degreesbelow the horizontal.

Part A

If the crate was at rest at the top of the incline and has a speed of 3.00 m/s at the bottom, find the change in the mechanical energy of the crate from the top to the bottom of the incline.

Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules.

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Part B

Find the change in the internal energy of the system from the initial to the final time.

Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules.

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Part C

Assume that all the internal energy change occurs in the crate itself. Find the change in temperature of the crate from the top of the incline to the bottom of the incline.

Conservation of Energy and Temperature Change

A crate of fruit with a mass of 39.5 kg and a specific heat of 3800 J/(kgK) slides 7.00 mdown a ramp inclined at an angle of 40.0 degreesbelow the horizontal.

Part A

If the crate was at rest at the top of the incline and has a speed of 3.00 m/s at the bottom, find the change in the mechanical energy of the crate from the top to the bottom of the incline.

Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules.

Emech   = J

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Part B

Find the change in the internal energy of the system from the initial to the final time.

Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules.

U   = J

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Part C

Assume that all the internal energy change occurs in the crate itself. Find the change in temperature of the crate from the top of the incline to the bottom of the incline.

T   = C

Explanation / Answer

Mechanical energy is the sum of potential and kinetic energy.

At the top of the incline, the crate only has potential energy.

At the bottom of the incline, it has only kinetic energy.

a) Mechanical energy at the top of the incline = potential energy at top of incline = mass x acc. due to gravity x vertical height of the crate = 39.5*9.81*7sin(40) = 1743.53 J

Mechanical energy at bottom of incline = kinetic energy at bottom of incline = .5 x mass x speed x speed = .5*39.5*3*3 = 177.75 J

Loss in mechanical energy = 1743.53-177.75= 1565.78 J

b) By conservation of energy, this loss of mechanical energy must be converted to increase of internal energy of the crate :

So, change in internal energy of the crate = 1565.78 J

c) It all the internal energy change occurs in the crate itself, the energy must raise the temperature of the crate. This change in temperature must be given by

T = (change in internal energy)/(mass x specific heat) = 1565.78/(39.5 * 3800) = 0.01 K

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