A 1.2 kg metal head of a geology hammer strikes a solid rock with a speed of 5.7 m/s. Assuming all the energy is retained by the hammer head, how much will it's temperature increase? C= .11 kcal/kgC
I already set up the problem. But I keep getting the wrong answer.
I have :
Temp change= 1/2 (1.2kg)(5.7m/s)^2 (1.00cal) divided by (1.2 kg)(.11 kcal/kgC) (4.18 J) • 1.00 kcal/1000cal
How do I solve this...
Nothing I come out with makes sense
To solve for the temperature change, you should use the equation:
Change in temperature = (1/2 * m * v^2) / (m * C * 4.18)
Let's break down the equation:
- "m" represents the mass of the hammer head, which is given as 1.2 kg.
- "v" represents the speed at which the hammer head strikes the rock, which is given as 5.7 m/s.
- "C" represents the specific heat capacity of the material, which is given as 0.11 kcal/kgC. However, it needs to be converted to J/kgC, so multiply by 4.18.
Now let's substitute the values into the equation:
Change in temperature = (1/2 * 1.2 kg * (5.7 m/s)^2) / (1.2 kg * 0.11 kcal/kgC * 4.18 J)
Simplifying further:
Change in temperature = (1/2 * 1.2 * 32.49) / (0.11 * 1.2 * 4.18)
Now calculate the numerator:
Change in temperature = 19.494 / (0.11 * 1.2 * 4.18)
Next, calculate the denominator:
Change in temperature = 19.494 / 0.552
Finally, solve for the change in temperature:
Change in temperature ≈ 35.3 °C
Therefore, the temperature of the hammer head will increase by approximately 35.3 degrees Celsius when it strikes the rock.