# Cooking chicken with a powerful slap

A question has been going around the social networks: If kinetic energy is converted to thermal energy upon impact, how hard to you need to slap a chicken to cook it? One response suggested raising the chicken’s temperature to 400 °F, which is far too hot, and it used an “average slap energy” that was not explained. Here then, is my analysis. You’re going to do what?!

Cooked chicken needs to be at least 165 °F (74 °C), and this calculation will assume a minimum temperature for ensuring safety. You can adjust to higher temperatures if you wish to subsequently sell the slapped chicken or if you are uncomfortable with that temperature.

The chicken is cooked starting at about 6 °C (just thawed). We are assuming perfect energy transfer, total conversion to thermal energy, with not even any sound produced by the slap.

The average weight of a broiler chicken is 2.84 kg (source). Average specific heat of a broiler is 3.22 kJ/(kg °C) (source). This implies a temperature change of 68 °C. The formula for specific heat is: $c = \frac{Q}{m \Delta T}$

where c = specific heat, Q = energy given, m = mass, and ΔT = change in temperature (in °C). If we rearrange it, we get: $Q = c m \Delta T$

So, (3.22 kJ/(kg °C))(2.84 kg)(68 °C) = 622 kJ or 622,000 Joules.

That’s a lot of energy. Let’s see how hard a slap would have to be.

The formula for force in Newtons is: $F = m a$

where m is mass in kg, and a is the acceleration, or change in velocity, in meters per second squared. Assuming a perfectly inelastic slap so that the hand stops dead, and all the kinetic energy of the slap is transferred into the thermal energy of the chicken, if v is the starting velocity in meters per second, we have the energy E: $E = \frac{1}{2} m v^2$

Let’s take the energy of a slap, then, to be the mass of the whole arm times the square of its velocity, all divided by 2. We can take the average weight of an adult arm to be about 2.5 kg. Then we have: $622 \text{ kJ} = \frac{1}{2} (2.5 \text{ kg}) v^2$

or $v = \sqrt{\frac{2 (622 \text{ kJ})}{2.5 \text{ kg}}} = 705 \frac{\text{m}}{\text{s}}$