The 1.19 kg head of a hammer reaches a speed of 8.4 m/s just before striking a nail and is brought to rest. Estimate the temperature increase of a 13 g iron nail after it is struck by twenty such hammer blows delivered in rapid succession, assuming the nail absorbs all the energy.

 First, let's calculate the kinetic energy of the hammer head just before it strikes the nail. The formula for kinetic energy (KE) is:

KE = (1/2) * m * v^2

where m is the mass of the hammer head and v is its velocity.

Plugging in the values, we have:

m = 1.19 kg v = 8.4 m/s

KE = (1/2) * 1.19 kg * (8.4 m/s)^2 KE = (1/2) * 1.19 kg * 70.56 m^2/s^2 KE = (1/2) * 1.19 kg * 70.56 J/kg*m^2/s^2 KE = 0.595 * 70.56 J KE ≈ 42 J (approx)

Now, we know that the hammer delivers 42 Joules of energy in one blow. When it strikes the nail twenty times, the total energy delivered is:

Total energy (E_total) = 42 J/hammer blow * 20 hammer blows E_total = 840 J

Now, we need to calculate the temperature increase for the iron nail. We can use the formula:

Q = mcΔT

where Q is the heat absorbed (which is the total energy E_total we just calculated), m is the mass of the nail, c is the specific heat capacity of iron, and ΔT is the change in temperature.

The mass (m) of the nail is 13 g, which is 0.013 kg. The specific heat capacity (c) of iron is roughly 450 J/kg·°C.

Using the energy absorbed by the nail (Q = E_total = 840 J) and rearranging the formula to solve for ΔT, we get:

ΔT = Q / (mc)

Substituting the values, we have:

ΔT = 840 J / (0.013 kg * 450 J/kg·°C) ΔT ≈ 840 J / (5.85 J/°C) ΔT ≈ 143.59 °C

Therefore, the estimated temperature increase of the iron nail after absorbing the energy from twenty hammer blows is approximately 143.59 °C.