A crate of fruit with a mass of 40.0 kg and a specific heat capacity of 3500 J/(

Question

A crate of fruit with a mass of 40.0 kg and a specific heat capacity of 3500 J/(kgK) slides 7.90 m down a ramp inclined at an angle of 36.3 degrees below the horizontal.

If the crate was at rest at the top of the incline and has a speed of 2.35 m/s at the bottom, how much work Wf was done on the crate by friction?

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

Answer:

Need help on part B

If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change T?

Explanation / Answer

Using the heat equation

Temperature variation symbol DT

Specific heat capacity symbol c

Q= c m DT

DT = Q/c m

Q=W = 1720 J

. c =3500 J / Kg K      

Variation of degree kelvin is equal to the change in degree Celsius

DT= 1720 / (3500 40)

DT= 0.012K = 0.012°C

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