120 J of heat flows by thermal conduction from 100$^{\circ}$C to 0.00$^{\circ}$C.What is the change in entropy for this process?
A) 0.034 J/K
B) 0.118 J/K
C) 0.201 J/K
D) 0.338 J/K

Question

Quizplus · Accepted Answer

We can use the formula for entropy change due to heat transfer by conduction:

ΔS = Q/T

where Q is the amount of heat transferred and T is the temperature at which the transfer takes place (in Kelvin).

First, we need to convert the temperatures given in Fahrenheit to Kelvin:

T1 = (100°F + 459.67) × 5/9 = 310.93 K
T2 = (0°F + 459.67) × 5/9 = 255.37 K

Substituting the values into the formula, we get:

ΔS = (120 J)/(310.93 K - 255.37 K) = 0.118 J/K

Therefore, the answer is B.

120 J of Heat Flows by Thermal Conduction from 100 ∘^{\circ}∘

120 J of Heat Flows by Thermal Conduction from 100 $^{\circ}$