Question

An apple with a mass of 0.12 kg and an average specific heat of 3.65 kJ/kg°C is cooled from 25°C to 5°C. Find the entropy change of the apple.

a. 1.152 kJ/°C
b. 0.576 kJ/°C
c. 2.304 kJ/°C
d. 1.728 kJ/°C

Answer 1

The change in entropy of the apple is approximately 0.876 kJ/°C.

Step-by-step explanation:

The change in entropy of an object can be calculated using the formula:

ΔS = mcΔT

Where:

• ΔS is the change in entropy
• m is the mass of the object
• c is the specific heat capacity of the object
• ΔT is the change in temperature

In this case, we have an apple with a mass of 0.12 kg and a specific heat capacity of 3.65 kJ/kg°C. The change in temperature is 25°C - 5°C = 20°C.

Plugging these values into the equation:

ΔS = (0.12 kg)(3.65 kJ/kg°C)(20°C) = 0.876 kJ/°C.

Therefore, the entropy change of the apple is approximately 0.876 kJ/°C.