Question

An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔSΔSDeltaS of the object? Assume that the temperature of the object does not change appreciably in the process

Answer 1

Answer:

Δ S = -50 J

Explanation:

given,

Temperature, T = 500 K

Heat dissipates, Q = 25.0 kJ

                            Q = 25000 J

change in entropy, ΔS = ?

using equation of entropy

 $\Delta S = \dfrac{\Delta Q}{T}$

 $\Delta S = -\dfrac{25000}{500}$

        Δ S = -50 J

negative sign is used because heat is lost in the surrounding.

Hence, Change in entropy is equal to -50 J