A piece of iron block moves across a rough horizontal surface before coming to rest. The mass of the block is 1.3 kg, and its initial speed is 4.0 m/s. How much does the block's temperature increase, if it absorbs 71% of its initial kinetic energy as internal energy? The specific heat of iron is 452 J / (kg · °C).

0.0126°C

Explanation:

Kinetic energy sustain by the block, K. E = 1/2 * mass*velocity^2

K. E = 1/2 * 1.3 * (4^2)

=1/2 * 1.3 * 16 = 1.3 * 8 = 10.4J

The internal energy is 71% of kinetic energy is;

U = 71/100 * 10.4 = 7.38J

But U = m * C * ∆T

Where U is internal energy

m is the mass of iron

C is the heat capacity of iron

∆T is the temperature change

By make ∆T the subject of the formula we have

∆T=U / m*C

=7.38 / 1.3*452 = 0.0126°C