A piece of iron block moves across a rough horizontal surface before coming to rest. The mass of the block is 1.8 kg, and its initial speed is 3.8 m/s. How much does the block's temperature increase, if it absorbs 69% of its initial kinetic energy as internal energy? The specific heat of iron is 452 J/(kg · °C).
=(1/2 m v^2)(.69)/(m C)
To determine how much the block's temperature increases, we need to calculate the amount of energy it absorbs as internal energy.
First, let's calculate the initial kinetic energy of the block using the formula:
Kinetic energy = (1/2) * mass * velocity^2
Substituting the given values:
Kinetic energy = (1/2) * 1.8 kg * (3.8 m/s)^2 = 12.95 J
Since the block absorbs 69% of its initial kinetic energy as internal energy, we can calculate the amount of energy absorbed as internal energy:
Internal energy absorbed = 0.69 * 12.95 J = 8.94 J
Now, we know that the internal energy gained is equal to the change in thermal energy, which is equal to the mass of the block multiplied by the specific heat capacity of iron multiplied by the change in temperature:
Internal energy gained = mass * specific heat * change in temperature
Rearranging the equation, we can solve for the change in temperature:
Change in temperature = Internal energy gained / (mass * specific heat)
Substituting the given values:
Change in temperature = 8.94 J / (1.8 kg * 452 J/(kg · °C)) ≈ 0.01 °C
Therefore, the block's temperature increases by approximately 0.01 °C.