
A ball is dropped on a floor from a height of 2m. After the collision it rises upto a height of 1m. Assuming that $20\% $ of mechanical energy is lost in the form of thermal energy. If the specific heat capacity of the ball is $800J/K$ then the rise in temperature of the ball during collision is (Take $g = 10m/{s^2}$).
A. $2.5 \times {10^{ - 3}}$ $^oC$
B. $7.5 \times {10^{ - 3}}$ $^oC$
C. $1.5 \times {10^{ - 3}}$ $^oC$
D. $5 \times {10^{ - 3}}$ $^oC$
Answer
170.4k+ views
Hint: As initially ball is at height of 2m and after collision ball height becomes 1m. So after collision loss in potential energy will increase the thermal energy of the ball i.e. decrease in potential energy will be the cause of increase in temperature. Hence change in potential energy will be equal to thermal energy.
Mathematically $mg({h_1} - {h_2}) = ms\Delta T$.
Complete step by step answer:
As we already know that change is potential energy will be equal to increase in thermal energy
$\therefore $given only 20% 0f mechanical energy will be convert into thermal energy
Hence, we apply the equation: $mg({h_1} - {h_2}) \times \dfrac{{20}}{{100}} = ms\Delta T$.
In the above equation m=mass of balL
g= acceleration due to gravity
${h_1}$=initial height
${h_2}$=final height
s=specific heat capacity
$\Delta T$=change in temperature of body
So after putting all given values in the above equation:
$
mg(2 - 1) \times \dfrac{{20}}{{100}} = m \times 800 \times \Delta T \\
\\
$ here $g = 10m{s^{ - 2}}$
$m \times 10 \times \dfrac{1}{5} = m \times 800 \times \Delta T$
$2 = 800 \times \Delta T$
$\Delta T = \dfrac{2}{{800}}$
$\Delta T = 2.5 \times {10^{ - 3}}$
Hence, increase in temperature of ball be $2.5 \times {10^{ - 3}}$
Option number A will be the correct option.
Note: Potential energy is the energy of the body associated with height of the body i.e. if a body will be more height from the surface of the earth then it will have more potential energy and if the body has more potential energy then it will have more mechanical energy.
Thermal energy is the energy of the body associated with the temperature of the body, thermal energy also depends on the specific heat of the body.
Mathematically $mg({h_1} - {h_2}) = ms\Delta T$.
Complete step by step answer:
As we already know that change is potential energy will be equal to increase in thermal energy
$\therefore $given only 20% 0f mechanical energy will be convert into thermal energy
Hence, we apply the equation: $mg({h_1} - {h_2}) \times \dfrac{{20}}{{100}} = ms\Delta T$.
In the above equation m=mass of balL
g= acceleration due to gravity
${h_1}$=initial height
${h_2}$=final height
s=specific heat capacity
$\Delta T$=change in temperature of body
So after putting all given values in the above equation:
$
mg(2 - 1) \times \dfrac{{20}}{{100}} = m \times 800 \times \Delta T \\
\\
$ here $g = 10m{s^{ - 2}}$
$m \times 10 \times \dfrac{1}{5} = m \times 800 \times \Delta T$
$2 = 800 \times \Delta T$
$\Delta T = \dfrac{2}{{800}}$
$\Delta T = 2.5 \times {10^{ - 3}}$
Hence, increase in temperature of ball be $2.5 \times {10^{ - 3}}$
Option number A will be the correct option.
Note: Potential energy is the energy of the body associated with height of the body i.e. if a body will be more height from the surface of the earth then it will have more potential energy and if the body has more potential energy then it will have more mechanical energy.
Thermal energy is the energy of the body associated with the temperature of the body, thermal energy also depends on the specific heat of the body.
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