Physics: A15-gram bullet moving at 1502 m/s plunges into 2.5 kg of paraffin wax. the wax was initially at 31°c. assuming that all the bullet s energy heats the wax, what is its final temperature (in ºc)? take the mechanical equivalent of heat to be 4 j/cal and the specific heat of wax to be 0.7 cal/g °c

A15-gram bullet moving at 1502 m/s plunges into

alondrabdiaz586

02.01.2022

33.4°C .

Explanation:

mass of bullet, m = 15 g = 0.015 kg

velocity of bullet, v = 1502 m/s

mass of wax, M = 2.5 kg

Initial temperature of wax, T1 = 31°C

Let T2 be the final temperature of wax.

Specific heat of wax, c = 0.7 cal/g°C = 0.7 x 1000 x 4 J/kg°C = 2800 J/kg°C

The kinetic energy of the bullet is converted into heat energy which is used to heat the wax.

$\frac{1}{2}mv^{2}= M \times c \times \left ( T_{2}-T_{1} \right )$

$0.5\times 0.015\times 1502 \times 1502 = 2.5 \times 2800 \times\left ( T_{2}-31 \right )$

$2.42 =\left ( T_{2}-31 \right )$

$T_{2}=33.4^{o}C$

thus, the final temperature of wax is 33.4°C .

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rimcdaniel2

28.12.2021

answer:

187m/s

explanation:

v = v0 + at

v = 115 + 2.88 * 25

v = 187m/s

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