Question

A football game begins by flipping a penny to decide which team will get the ball first. The 5.25 g penny has a speed of 3.27 m/s just before it hits the ground. If half of the increase in the internal energy of the ground and penny is transferred to the penny’s internal energy, by how much does the penny’s temperature increase after impact? Assume that the penny’s temperature increases by 1.00° C as 2.03 J are added to its mass.

Answer 1

Answer:

$2.6 ^{\circ}C$

Explanation:

First of all, we need to calculate the penny's kinetic energy before hitting the ground. This is given by

$K=\frac{1}{2}mv^2$

where m = 5.25 g = 0.00525 kg is the penny's mass and v = 3.27 m/s is its speed. Substituting,

$K=\frac{1}{2}(0.00525 kg)(3.27 m/s)^2=0.028 J$

When the penny hits the ground, all this energy is converted into internal energy of the penny and the ground. If only half is converted into penny's internal energy, its increase in internal energy is

$\Delta U= \frac{0.028 J}{2}=0.014 J$

And its formula is

$\Delta U=m C_s \Delta T$

where m is the penny's mass, Cs its specific heat capacity (2.03 J/gC) and $\Delta T$ the increase in temperature. Solving for the last term,

$\Delta T=\frac{\Delta U}{m C_s}=\frac{0.028 J}{(0.00525 kg)(2.03 J/gC)}=2.6 ^{\circ}C$