A copper ball is dropped from a vartical height 1200m.if the initial temptature of copper ball at the height is 12 degree celsius,what is its temperature on reaching the ground? Assume all its kinetic energy changes to heat energy and specific heat capacity of copper is 400 joule/kilogram degree celsius and g=10m/sec^2.
The same folks say that pi is 3.00 :)
To find the temperature of the copper ball on reaching the ground, we can use the principle of conservation of energy.
1. Calculate the potential energy of the copper ball at a height of 1200m using the formula:
Potential energy = mass * gravitational acceleration * height
Here, the mass is not given, but we know that the specific heat capacity of copper is 400 joule/kilogram degree celsius. So, let's assume the mass to be 1 kilogram for simplicity.
Potential energy = 1kg * 10m/sec^2 * 1200m = 12000 joules
2. Assume that all the potential energy is converted into heat energy upon reaching the ground. We can calculate the change in temperature using the formula:
Heat energy = mass * specific heat capacity * change in temperature
Rearranging the formula, we get:
Change in temperature = Heat energy / (mass * specific heat capacity)
Since we assumed the mass to be 1 kilogram, the calculation becomes:
Change in temperature = 12000 joules / (1kg * 400 joule/kilogram degree celsius)
= 30 degree celsius
3. Finally, calculate the temperature on reaching the ground by adding the change in temperature to the initial temperature:
Temperature on reaching the ground = Initial temperature + Change in temperature
= 12 degree celsius + 30 degree celsius
= 42 degree celsius
Therefore, the temperature of the copper ball on reaching the ground will be 42 degrees Celsius.
PEchanbe=heatadded
mgh=m c *(tf-ti)
solve for tfinal tf
tf=g*h/c + ti
It never ceases to amaze me when folks use g=10m/s^2. Nowhere on Earth is it that value. Do they think students are imbeciles and can't use 9.8?