a 1.0 kg metal head of a geology hammer strikes a solid rock with a speed of 3.9 m/s. assuming all the energy was retained by the hammer head, how much will its temperature increase?
masshammer*specheat*deltaTemp=1/2*mass*3.9^2
7.605
To determine the increase in temperature of the hammer head, we can use the principle of conservation of energy. The energy gained by the hammer head is equal to the work done on it.
First, let's calculate the kinetic energy of the hammer head. Kinetic energy is given by the formula:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
In this case, the mass of the hammer head is 1.0 kg and the velocity is 3.9 m/s. Plugging these values into the formula, we get:
KE = (1/2) * 1.0 kg * (3.9 m/s)^2
= 0.5 * 1.0 kg * 15.21 m^2/s^2
= 7.605 Joules
Now, let's assume all the energy is retained by the hammer head and converted into thermal energy. To find out how much the temperature increases, we need to determine the heat capacity of the hammer head. The heat capacity is a property that quantifies how much energy is required to raise the temperature of an object by a certain amount.
The specific heat capacity (c) is defined as the amount of heat (q) required to raise the temperature (ΔT) of a unit mass (m) of a substance by one degree Celsius (or Kelvin). Mathematically, it can be expressed as:
c = q / (m * ΔT)
Rearranging the formula, we can find the change in temperature:
ΔT = q / (m * c)
In this case, the heat (q) is equal to the kinetic energy (KE) gained by the hammer head, the mass (m) is 1.0 kg, and we need to find the change in temperature (ΔT).
Let's assume that the specific heat capacity of the hammer head is 450 J/(kg·°C) (a rough approximation for metals). Plugging in the values, we have:
ΔT = 7.605 J / (1.0 kg * 450 J/(kg·°C))
≈ 0.0169 °C
Therefore, the temperature of the hammer head would increase by approximately 0.0169 degrees Celsius.