Question

A crate of fruit with a mass of 35.0 kg and a specific heat capacity of 3800 J/(kg⋅K) slides 8.40 m down a ramp inclined at an angle of 36.1 degrees below the horizontal.

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 ΔT?

Answer 1

given

m = 35 kg ,

C = 3800 J/(kg⋅K) ,

L = 8.40 m ,

= 36.1 degrees ,

v = 1 m/sec ( velocity not mentioned )

using h = L cos

h = 8.4 X cos36.1

h = 6.78 m

PE₁ = m g h and PE₂ = 0

and KE₁ = 0 and KE₂ = 1/2 mv²

TE₁ = PE₁ + KE₁

TE₁ = m g h

same way TE₂ = 1/2 mv²  

TE = TE₁ - TE₂

TE = m g h - 1/2 m v²   

TE = 35 X 9.8 X 6.78 - 0.5 X 35 X v²  

TE = 2325.54 - 17.5

TE = 2308.04 J

Q = m C T

T = Q / m C

T = 2308.04 / 35 X 3800

T = 0.0173^oC

the temperature change T = 0.0173^oC