he temperature of a cold orange with an average mass of 0.28 kg and average specific heat of 3.42 kJ/kg-degC rises from 7 degrees C to 25 degrees C as a result of heat transfer from the surounding air. The entropy change of the orange isQuestion 3Select one:a.0.213 kJ/Kb.1.22 kJ/Kc.0.0597 kJ/Kd.0 kJ/Ke.0.154 kJ/K

The entropy change of a system can be calculated using the formula:

ΔS = ∫dQ/T

where ΔS is the change in entropy, dQ is the infinitesimal amount of heat transferred to the system, and T is the absolute temperature.

However, for a process where the temperature change is relatively small (like in this case), we can simplify this to:

ΔS = Q/T

where Q is the total heat transferred to the system.

First, we need to calculate the total heat transferred to the orange. We can do this using the formula:

Q = mcΔT

where m is the mass of the orange, c is the specific heat of the orange, and ΔT is the change in temperature.

Substituting the given values:

Q = (0.28 kg)(3.42 kJ/kg-degC)(25 degC - 7 degC) = 17.352 kJ

Next, we need to convert this to Joules (since the units of entropy are Joules per Kelvin, not kilojoules per Kelvin):

Q = 17.352 kJ * 1000 J/kJ = 17352 J

Next, we calculate the average temperature during the process in Kelvin. This is simply the average of the initial and final temperatures, converted to Kelvin:

T_avg = ((7 degC + 25 degC)/2) + 273.15 K = 302.15 K

Finally, we can calculate the entropy change:

ΔS = Q/T_avg = 17352 J / 302.15 K = 0.0574 kJ/K

So, the closest answer to this value is c. 0.0597 kJ/K.