A 0.10 kg of steel ball falls on earth from a height of 10 m and bounces to a height of 1.0m . Let all the dissipated energy were absorbed by the ball as heat, calculate the rise in temperature if specific heat capacity of is 0.12 cal/g/deg C.
energy from falling: mgh=.1*9.8*10
energy to get to 1m= mgh=.1*9.8*1
subtract those, and you get heat energy.
Heatenergy=m*c*Changtemp
solve for changeIntemp
To calculate the rise in temperature of the steel ball, we need to determine the energy dissipated during the fall and bounce, and then convert that energy into an increase in temperature using the specific heat capacity.
1. Find the potential energy of the steel ball at the initial height:
Potential energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)
Given: mass (m) = 0.10 kg, height (h) = 10 m, and acceleration due to gravity (g) = 9.8 m/s^2
PE = 0.10 kg * 9.8 m/s^2 * 10 m = 9.8 J
2. Calculate the total energy dissipated during the fall and bounce:
The total energy dissipated is equal to the initial potential energy minus the final potential energy.
The final potential energy can be calculated using the mass (m), acceleration due to gravity (g), and the new height (h') after bouncing.
Given: height after bouncing (h') = 1.0 m
Final potential energy (PE') = 0.10 kg * 9.8 m/s^2 * 1.0 m = 0.98 J
Energy dissipated = Initial potential energy - Final potential energy
= 9.8 J - 0.98 J = 8.82 J
3. Convert the energy dissipated into a temperature rise in Celsius using the specific heat capacity of steel and the mass of the ball:
The energy dissipated is absorbed by the ball as heat, and we can calculate the temperature rise using the equation:
Energy dissipated (Q) = mass (m) * specific heat capacity (c) * temperature rise (ΔT)
Given: specific heat capacity (c) = 0.12 cal/g/°C
1 g = 0.001 kg (conversion factor)
Rearrange the equation to solve for the temperature rise (ΔT):
ΔT = Q / (m * c)
Since the specific heat capacity is given in cal/g/°C, we need to convert it to J/kg/°C by multiplying it by 4.18 (conversion factor).
c = 0.12 cal/g/°C * 4.18 J/cal = 0.5016 J/g/°C
ΔT = 8.82 J / (0.10 kg * 0.5016 J/g/°C)
= 175.1 °C
Therefore, the rise in temperature of the steel ball is approximately 175.1 °C.