A block of mass m = 0.2 kg slides across a rough horizontal surface with coefficient of kinetic friction $\mu$_k = 0.5.What is the change in entropy after the block has moved a distance of 1 m? The temperature of the block and surrounding is 22 $\circ$ C.

Question

Quizplus · Accepted Answer

The change in entropy can be calculated using the formula ∆S = Q/T, where Q is the heat energy transferred and T is the temperature. In this case, the block is losing kinetic energy due to friction, so the heat energy transferred can be calculated as Q = -Ffriction * distance, where Ffriction is the force of friction and distance is the distance travelled. The force of friction is given by Ffriction = µmg, where µ is the coefficient of kinetic friction, m is the mass of the block, and g is the acceleration due to gravity. Putting these values in the equation, we get:

Q = -µmg * distance = -0.5 * 0.2 kg * 9.81 m/s^2 * 1 m = -0.981 J

The temperature is 22°C, which is equivalent to 295 K. Therefore, the change in entropy is:

∆S = Q/T = -0.981 J / 295 K = -3.33 * 10^-3 J/K

We take the absolute value and get 3.3J/K. So, the best choice is B.

A Block of Mass M = 0 μ\muμ k = 0 ∘\circ∘

A Block of Mass M = 0 $\mu$ _k = 0 $\circ$