a 1.0 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, how much will its temperature increase? (Chead=0.11 kcal/kg degrees C)

1/2 m v^2 = mc Deltatemp
solve for deltaTemp.

0.00132

To solve for the change in temperature (ΔT), we can use the formula:

ΔT = (1/2) (mv^2) / (mC_head)

Given:

Head mass (m) = 1.0 kg
Velocity (v) = 5.0 m/s
Specific heat of the head material (C_head) = 0.11 kcal/kg °C

Let's substitute the values into the formula:

ΔT = (1/2) (1.0 kg) (5.0 m/s)^2 / (1.0 kg) (0.11 kcal/kg °C)

First, let's simplify the expression:

ΔT = (1/2) (1.0) (25 m^2/s^2) / (1.0) (0.11 kcal / °C)

ΔT = 12.5 m^2/s^2 / (0.11 kcal / °C)

Next, we need to convert the unit of the result from kcal to degrees Celsius. We know that 1 calorie (cal) is equivalent to 0.001 kcal, and 1 cal is equivalent to a temperature change of 1°C for 1 gram of water. Since we are dealing with kilograms, we need to multiply the result by 1000:

ΔT = (12.5 m^2/s^2 / (0.11 kcal / °C)) * 1000 cal/kcal

ΔT = (12.5 m^2/s^2 / 0.11 kcal / °C) * 1000 cal/kcal

Now, we can cancel out the kcal units:

ΔT = (12.5 m^2/s^2) * (1000 cal/kcal) / (0.11 °C)

ΔT = 113,636.36 cal / °C

Therefore, the change in temperature (ΔT) of the metal head of the geology hammer will be approximately 113,636.36 °C.