A 25-kg iron block initially at 350oC is quenched in an insulated tank that c ontains 100 kg of water at 18oC. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process.

4.08kg/J

Explanation: Please find the attached file for the solution

Entropy = 4.08 kj/k

Explanation:

From energy balance in first law of thermodynamics, we have;

Δv(i)+ ΔU(h2o) = 0

Thus;

[MCp(T2 - T1)]iron + [MCp(T2 - T1)]water = 0

Where Cp is specific heat capacity

For iron, Cp = 0.45 Kj/kg°C and for water, Cp = 4.18 Kj/kg°C

From question, Mass of iron =25kg while mass of water = 100kg

And Initial temperature of iron (T1) = 350°C while initial temperature of water(T1) = 18°C

Thus,

[25 x 0.45(T2 - 350)] + [100 x 4.18(T2 - 18)] = 0

11.25T2 - 3937.5 + 418T2 - 7524 = 0

So,

429.25T2 = 11461.5

T2 = 26.7 °C

Now for entropy, we have convert the temperature from degree celsius to kelvins.

Thus, for iron T1 = 350 + 273 = 623K and for water, T1 = 18 + 273 = 291 K. Also, T2 = 26.7 + 273 = 299.7K.

The entropy changes will be;

For iron ;

Δs(i) = MCp(In(T2/T1)) = 25 x 0.45(In(299.7/623)) = -8.23 Kj/k

Now, for water;

Δs(water) = MCp(In(T2/T1)) = 100 x 4.18(In(299.7/291)) = 12.31 kj/k

Thus, total entropy will be the sum of that of iron and water.

Δs(total) = 12.31 kj/k - 8.23 Kj/k = 4.08 kj/k