An iron balls is drop into a congrats floor from a height of 10m. On the first rebouce it rises to 0.5m. Assume that all the macroscopic mechanical energy lost in cotillion with the floor goes into the ball and that specific heat capacity of is 501.6J/kg°c during the collision. Has heat being added to the ball?. Has work done on it? Has internal energy changed? If so, by how much?. How much have temperature of the ball risen after the first coillison?
To determine whether heat has been added to the ball, work has been done on it, and if there has been a change in internal energy, we need to analyze the situation.
- Heat: Heat is the transfer of energy due to a temperature difference. Since there is no mention of any external heat source or temperature difference in the problem, we can conclude that no heat has been added to the ball.
- Work: Work is the transfer of energy due to the application of force over a distance. In this case, when the ball hits the floor and rebounds, work is done on the ball as it moves against the force of gravity. The work done can be calculated using the formula: work = force × distance. Since the force acting on the ball is its weight, which is given by the equation: weight = mass × gravity, we can calculate the work done.
- Internal Energy: Internal energy is the sum of the kinetic and potential energies of the particles within a system. When the ball hits the floor, it loses some of its mechanical energy as it deforms and experiences a change in velocity. This lost mechanical energy is converted into other forms, such as thermal energy and sound. Therefore, there is a change in internal energy.
To calculate the amount of work done on the ball, we need to determine the force acting on it. Using the equation for weight, weight = mass × gravity, we can calculate the force. The mass of the ball is not given in the problem, so we assume a value. Let's assume the mass of the ball is 1 kg.
Weight = mass × gravity = 1 kg × 9.8 m/s^2 = 9.8 N
The work done is equal to the force multiplied by the distance. In this case, the distance is the height from which the ball is dropped (10 m). Therefore, the work done on the ball is:
Work = force × distance = 9.8 N × 10 m = 98 J
The internal energy changes as some of the mechanical energy is converted into other forms, such as thermal energy and sound. Since there is a loss in mechanical energy and an increase in internal energy, we can conclude that the internal energy of the ball has increased.
To calculate the change in temperature, we need to know the specific heat capacity of the material. According to the problem, the specific heat capacity of the ball is 501.6 J/kg°C. However, we need the mass of the ball to find the change in temperature. Assuming a mass of 1 kg, we can proceed with the calculation.
Change in temperature = (Work done on the ball) / (Mass × Specific heat capacity)
= 98 J / (1 kg × 501.6 J/kg°C)
≈ 0.1956°C
Therefore, after the first collision, the temperature of the ball has risen by approximately 0.1956°C.