A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained by the hammer head, and that it's heat capacity is 0.11 cal/(gC), how much will its temperature increase?
Well, before we start, let's just address the "hot rocks" situation. I understand that rocks can be "solid," but I don't think they're big fans of your hammer's advances. They're pretty resistant, you know? It's probably better to stick to a rock concert with your hammer skills.
Now, coming back to your question, we can calculate the change in temperature using the formula:
ΔQ = m * C * ΔT
Where:
ΔQ is the change in heat energy (which in this case is equal to the kinetic energy)
m is the mass of the hammer head (1.0 kg)
C is the heat capacity of the hammer head (0.11 cal/(g°C))
ΔT is the change in temperature that we're trying to find.
Since the velocity of the hammer head is given, you can calculate the kinetic energy using this formula:
KE = (1/2) * m * v^2
Once you find the kinetic energy, you can set ΔQ equal to that value and solve for ΔT. And remember, always stay cool, even when dealing with hot rocks!
To calculate the temperature increase of the geology hammer head, we can use the specific heat formula:
Q = mcΔT
Where:
Q = heat energy (in calories)
m = mass of the hammer head (in grams)
c = specific heat capacity (in calories/(gram°C))
ΔT = change in temperature (in °C)
First, let's convert the mass of the hammer head from kilograms to grams:
m = 1.0 kg x 1000 g/kg = 1000 g
Next, we can calculate the heat energy by using the kinetic energy formula:
KE = 0.5mv^2
Where:
m = mass of the hammer head (in kg)
v = velocity of the hammer head (in m/s)
KE = 0.5 x 1.0 kg x (5.0 m/s)^2
KE = 0.5 x 1.0 kg x 25.0 m^2/s^2
KE = 12.5 J
Since 1 cal = 4.184 J, we can convert the energy to calories:
Q = 12.5 J / 4.184 J/cal
Q = 2.988 cal
Finally, we can calculate the temperature change using the specific heat formula:
2.988 cal = (1000 g) x (0.11 cal/(g°C)) x ΔT
ΔT = 2.988 cal / (1000 g x 0.11 cal/(g°C))
ΔT = 2.988 / 110
ΔT ≈ 0.027 °C
Therefore, the temperature of the hammer head is expected to increase by approximately 0.027 °C.
To calculate the change in temperature of the metal head of the geology hammer, we can use the formula:
ΔQ = m * c * ΔT
Where:
ΔQ is the change in heat energy
m is the mass of the metal head (in grams)
c is the heat capacity of the metal head (in cal/(g⋅°C))
ΔT is the change in temperature (in °C)
First, we need to convert the mass of the metal head from kilograms to grams:
mass of the metal head = 1.0 kg * 1000 g/kg = 1000 g
Now, let's calculate the change in heat energy:
ΔQ = m * c * ΔT
ΔQ = 1000 g * 0.11 cal/(g⋅°C) * ΔT
We know that all the energy is retained by the hammer head, so we can equate the change in heat energy to the initial kinetic energy of the hammer head:
ΔQ = (1/2) * m * v^2
Where:
m is the mass of the hammer head (in kg)
v is the velocity of the hammer head (in m/s)
Plugging in the values, we get:
1000 g * 0.11 cal/(g⋅°C) * ΔT = (1/2) * 1.0 kg * (5.0 m/s)^2
Now, let's solve for ΔT:
1000 g * 0.11 cal/(g⋅°C) * ΔT = 12.5 J
ΔT = 12.5 J / (1000 g * 0.11 cal/(g⋅°C))
To convert the energy from joules to calories, we need to divide by 4.18:
ΔT = (12.5 J / (1000 g * 0.11 cal/(g⋅°C))) / 4.18
Calculating this expression gives us the change in temperature:
ΔT = 2.98 °C
Therefore, the temperature of the metal head of the geology hammer will increase by 2.98°C.