If the kinetic energy is converted into thermal energy, how hard do I have to slap a pork belly to cook it?

Someone asked me that “if the kinetic energy is converted into thermal energy, how hard do I have to slap a pork belly to cook it?”

As his friendly colleague a Mechanical Engineer, I decided to calculate this with a few assumptions.

The formula for converting between Kinetic Energy and Thermal Energy is 1/2Mv^2 = mc(T-t)
Where:
M=average weigh of human hand
m=average weigh of pork belly
v=average velocity human slap
c=specific heat constant of pork belly
T=temperature of meat for us to consider it cooked
t=ambience temperature or temperature of meat before impact.

M= 0.4kg, average weigh of human hand
m=6kg, average weigh of pork belly
v=11m/s
c= 1590 J/kg.K,considering 60% water content. (https://www.sciencedirect.com/topics/food-science/thermal-property-of-food)
T=322°C, for us to consider it cooked.
t=20°C, room temperature.

Computing the number of slaps to cook the pork belly;

Let n be the number of slaps
1/2Mv^2(n)=mc(T-t)
½(0.4kg)(11m/s)^2(n)=6kg(1590J/kg.K)(322-20)K
24.2J(n)= 6kg(1590J/kg.K)(322-20)K
24.2J(n)= 2,881,080J
[24.2J(n)} / 24.2J = (2,881,080J) / 24.2J
[24.2J(n)} / 24.2J = (2,881,080J) / 24.2J
n=119,053
Hence, it would take 119,053 average slaps to cook the pork belly.

Computing the velocity to cook the pork bell with one slap;

1/2Mv^2=mc(T-t)
½(0.4kg)(v^2)=(6kg)(1590J/kg.K)(322-20)K
0.2kg(v^2)= 2,881,080J
J=Joules, J=kN.m=kg.m^2/s2
[0.2kg(v^2)]/ 0.2kg = (2,881,080 kg.m^2/s2)/ 0.2kg
v^{2=14,405,400m}2/s^2
taking the square root of both side;
v=3,796 m/s
To cook the pork belly in one slap, you would have to slap it with a velocity of 3,796 m/s or 8491.41mph.