A 9.0 kg iron ball is dropped onto a pavement from a height of 110 m.If half of the heat generated goes into warming the ball, find the temperature increase of the ball. (In SI units, the specific heat capacity of iron is 450 J/kg*degree C.)
Set 1/2(M*g*H) equal to M*C*deltaT
amd solve for deltaT, the temperature change of the iron. Note that the mass M cancels out. The specific heat (C) has been provided to you.
H = 110 m
g = 9.81 m/s^2
deltaT = g*H/(2*C)
To find the temperature increase of the iron ball, we need to calculate the heat generated and then use the specific heat capacity of iron.
Step 1: Calculate the potential energy (PE) of the ball.
The potential energy can be calculated using the formula: PE = m * g * h, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the ball is dropped.
PE = 9.0 kg * 9.8 m/s^2 * 110 m
PE = 9.0 kg * 1078 N
PE = 9702 J
Step 2: Calculate the heat generated.
Since half of the heat generated goes into warming the ball, we need to divide the potential energy by 2 to get the heat generated.
Heat generated = PE / 2
Heat generated = 9702 J / 2
Heat generated = 4851 J
Step 3: Calculate the temperature increase.
The formula to calculate the heat (Q) absorbed or released by an object is: Q = m * c * ΔT,
where Q is the heat absorbed or released, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
Since we want to find the temperature increase, we rearrange the formula to ΔT = Q / (m * c).
ΔT = 4851 J / (9.0 kg * 450 J/kg*°C)
ΔT = 4851 J / 4050 J/°C
ΔT = 1.2 °C
Therefore, the temperature of the iron ball will increase by 1.2 degrees Celsius.