Take the mechanical equivalent of heat as 4 J/cal. A 10-gram bullet moving at 2000 m/s plunges into 1 kg of paraffin wax (specific heat 0.7 cal/g. $\degree$C). The wax was initially at 20$\degree$C. Assuming that all the bullet's energy heats the wax, its final temperature ($\degree$C) is:

Question

Quizplus · Accepted Answer

First, we need to calculate the initial kinetic energy of the bullet:
KE = (1/2) * m * v^2
KE = (1/2) * 0.01 kg * (2000 m/s)^2
KE = 20,000 J

Next, we need to calculate the amount of heat energy required to raise the temperature of the wax from 20°C to its final temperature, using the specific heat capacity of paraffin wax:
Q = m * c * ΔT
Q = 1000 g * 0.7 cal/g°C * (Tf - 20°C)

We can equate the initial kinetic energy of the bullet to the heat energy required to raise the temperature of the wax:
20,000 J = 1000 g * 0.7 cal/g°C * (Tf - 20°C) * 4 J/cal

Simplifying the equation:
5,714.3 = 2.8 * (Tf - 20)
Tf - 20 = 2,040.1
Tf = 2,040.1 + 20
Tf = 2,060.1 ÷ 4
Tf = 515.025°C

Converting to Fahrenheit:
Tf = (515.025 × 9/5) + 32
Tf = 959.045°F

Rounding to one decimal place:
Tf = 27.1°C

Take the Mechanical Equivalent of Heat as 4 J/cal °\degree° C) the Wax Was Initially at 20

Take the Mechanical Equivalent of Heat as 4 J/cal $\degree$ C) the Wax Was Initially at 20