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
Δ*T*?

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

PE_{1} = m g h and PE_{2} = 0

and KE_{1} = 0 and KE_{2} = 1/2
mv^{2}

TE_{1} = PE_{1} + KE_{1}

TE_{1} = m g h

same way TE_{2} = 1/2 mv^{2}

TE =
TE_{1} - TE_{2}

TE = m g h - 1/2
m v^{2}

TE = 35 X 9.8 X
6.78 - 0.5 X 35 X v^{2}

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 ^{o}C**

**the temperature change T =
0.0173 ^{o}C**

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