Question

A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J K−1.

Answer 1

the answer to your questions is

Answer 2

The rise temperature of the ball in the collision is $2.45\times10^{-3}^{\circ}C$

Explanation:

Given that,

Initial height = 2.0 m

Final height = 1.5 m

Heat capacity = 800 J/K

Since potential energy = mechanical energy for a body at rest

As kinetic energy is zero.

We need to calculate the mechanical energy lost

Using formula of mechanical energy

$\text{mechanical energy lost}=mg(h_{1}-h_{2})$

Put the value into the formula

$\text{mechanical energy lost}=m9.8(2-1.5)$

$\text{mechanical energy lost}=4.9m$

We need to calculate the rise temperature

Using formula of heat

$\text{40\% of mechanical energy lost}=mc\Delta T$

Put the value into the formula

$\dfrac{40}{100}\times4.9m=m\times800\times\Delta T$

$\Delta T=\dfrac{40\times4.9}{100\times800}$

$\Delta T=2.45\times10^{-3}^{\circ}C$

Hence, The rise temperature of the ball in the collision is $2.45\times10^{-3}^{\circ}C$

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Topic : rise temperature

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