Question

A 45-kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18°C. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process. The specific heat of water at 25°C is cp = 4.18 kJ/kg·K. The specific heat of iron at room temperature is cp = 0.45 kJ/kg·K

Answer 1

Answer: −3.46kJ/K

Explanation:

From the question above, we have:

The mass of the block (m) = 45kg

The initial temperature of the block (T1) = 280∘C

The weight of the water (mw) = 100kg

The temperature of water (Tw) = 18∘C

Recall the energy balance equation,

ΔUI = −ΔUw

In this case ΔUI is the internal energy of the iron, while ΔUw is the internal energy of water.

[mcp (T2 − T1)]I = −[mcp (T2 − T1)]w

Here cp is the specific heat at constant pressure.

The specific heat of iron is (cp)I = 0.45kJ/kg⋅K, and the specific heat of water is (cp)w = 4.18kJ/kg⋅K.

Now, we substitute the values in above equation,

[45 × 0.45(T2 − 280)]I = −[100 × 4.18(T2 − 18)]w

[20.25(T2 − 280)] = −[418(T2 − 18)]

20.25T2 − 5,670 = −[418T2 − 7,524]

20.25T2 − 5,670 = −418T2 + 7,524

20.25T2 + 418T2 = 7,524 + 5,670

438.35T2 = 13,194

T2 = 30.1K

Recall, the expression to calculate the total entropy change is given as:

ΔStotal = ΔSI + ΔSw

ΔStotal = [mcpln(T2/T1)]I + [mcpln(T2/T1)]w

Now, we substitute the values in above equation,

ΔStotal = [45 × 0.45ln(297.6/553)]I + [100 × 4.18ln(297.6/291)]w

ΔStotal = −12.55 + 9.09

ΔStotal = −3.46kJ/K

Thus the total entropy change is −3.46kJ/K.