A 2.00-kg block of ice at 0.00°C is dropped into a very large lake at 25.0°C and completely melts.For water,the heat of fusion is 3.35 × 10⁵ J/kg,the heat of vaporization is 2.25 × 10⁵ J/kg,and the specific heat is 4190 J/kg ∙ K.The net change in entropy of the system consisting of the ice and the lake due to this melting process is closest to A)2.45 × 10³ J/K. B)2.24 × 10³ J/K. C)2.06 × 10² J/K. D)-2.45 × 10³ J/K. E)-2.06 × 10² J/K.

Question

Quizplus · Accepted Answer

The melting of the ice can be divided into two steps:
Step 1: Heating the ice from 0°C to 0°C with no phase change.
Step 2: Melting the ice at 0°C to water at 0°C.
For step 1, we can use the specific heat of ice:
Q = cmΔT = (2.00 kg)(2090 J/kg ∙ K)(0-(-25)) = 104500 J
For step 2, we need to calculate the heat required for melting the ice:
Q = mL = (2.00 kg)(3.35 × 10^5 J/kg) = 670000 J
The total heat required for the process is:
Q_total = 104500 J + 670000 J = 774500 J
The change in entropy of the system can be calculated using the formula:
ΔS = Q_total/T = 774500 J/(273+25) K = 2061.7 J/K
We can round this to two significant figures to get the answer: 2.06 × 10^12 J/K. Therefore, the best choice is C.

A 200-Kg Block of Ice at 0