
A bullet moving with velocity v collides against a wall consequently half of its kinetic energy is converted into heat. If the whole heat is acquired by the bullet the rise in temperature will be-
A. $\dfrac{{{v}^{2}}}{4s}$
B. $\dfrac{4{{v}^{2}}}{2s}$
C. $\dfrac{{{v}^{2}}}{2s}$
D. $\dfrac{{{v}^{2}}}{s}$
Answer
553.8k+ views
Hint: As a first step, you could recall the expression for kinetic energy. Then you could find the heat absorbed by the bullet which would be half of the kinetic energy. Then you could recall the expression for heat absorbed in terms of specific heat and then equate it to the above expression and hence find the answer.
Formula used:
Kinetic energy,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
Heat absorbed,
$Q=ms\Delta T$
Complete solution:
In the question, we have a bullet that is moving with a velocity v. This bullet is seen to collide against a wall. Due to this collision, half of the bullet’s kinetic energy is being converted into heat. We are asked to find the rise in temperature under the condition that the whole heat is acquired by the bullet.
Let us recall the expression for kinetic energy that is given by the expression,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
We are said that half of this kinetic energy is transformed into heat energy. Therefore, heat absorbed by the bullet can be given by,
$Q=\dfrac{1}{2}\left( \dfrac{1}{2}m{{v}^{2}} \right)=\dfrac{1}{4}m{{v}^{2}}$ …………………………………. (1)
Now let us recall the expression for heat absorbed in terms of specific heat. So, the expression can be given by,
$Q=ms\Delta T$ ……………………………………….. (2)
Where s is the specific heat and $\Delta T$ is the change in temperature.
Equating equations (1) and (2), we get,
$\dfrac{1}{4}m{{v}^{2}}=ms\Delta T$
$\therefore \Delta T=\dfrac{{{v}^{2}}}{4s}$
Therefore, we found the rise in temperature to be given by,
$\Delta T=\dfrac{{{v}^{2}}}{4s}$
Hence, option A is found to be the correct answer.
Note:
Heat capacity of a substance can be defined as the heat absorbed by a substance for the change in temperature by $1{}^\circ $. Specific heat capacity of a substance could be defined as the heat absorbed per unit gram of substance to result in a temperature change by $1{}^\circ $. This change in temperature can be increased or decreased depending on whether the substance has absorbed or emitted heat energy.
Formula used:
Kinetic energy,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
Heat absorbed,
$Q=ms\Delta T$
Complete solution:
In the question, we have a bullet that is moving with a velocity v. This bullet is seen to collide against a wall. Due to this collision, half of the bullet’s kinetic energy is being converted into heat. We are asked to find the rise in temperature under the condition that the whole heat is acquired by the bullet.
Let us recall the expression for kinetic energy that is given by the expression,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
We are said that half of this kinetic energy is transformed into heat energy. Therefore, heat absorbed by the bullet can be given by,
$Q=\dfrac{1}{2}\left( \dfrac{1}{2}m{{v}^{2}} \right)=\dfrac{1}{4}m{{v}^{2}}$ …………………………………. (1)
Now let us recall the expression for heat absorbed in terms of specific heat. So, the expression can be given by,
$Q=ms\Delta T$ ……………………………………….. (2)
Where s is the specific heat and $\Delta T$ is the change in temperature.
Equating equations (1) and (2), we get,
$\dfrac{1}{4}m{{v}^{2}}=ms\Delta T$
$\therefore \Delta T=\dfrac{{{v}^{2}}}{4s}$
Therefore, we found the rise in temperature to be given by,
$\Delta T=\dfrac{{{v}^{2}}}{4s}$
Hence, option A is found to be the correct answer.
Note:
Heat capacity of a substance can be defined as the heat absorbed by a substance for the change in temperature by $1{}^\circ $. Specific heat capacity of a substance could be defined as the heat absorbed per unit gram of substance to result in a temperature change by $1{}^\circ $. This change in temperature can be increased or decreased depending on whether the substance has absorbed or emitted heat energy.
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