An isolated system consist of a 10 kg copper slab initially

An isolated system consist of a 10 kg copper slab, initially at 30 degree C, and 0.2 kg of saturated water vapor, initially at 130 degree C. Assuming no volume change, determine the final equilibrium temperature of the isolated system, in degree C.

Solution

Specific heat of copper C1 = 0.385 kJ/kg-K

Average temperature of vapor = (100+130)/2 = 115 deg C

At 115 deg C, Specific heat of vapor C2 = 1.895 kJ/kg-K

Latent heat of evaporation for water L = 2260 kJ/kg

Specific heat of water C3 = 4.186 kJ/kg-K

Assume T < 100 deg C

The copper slab will get heat Q1 = C1 * m1 * (T - 30), where 1 - specific heat of copper, m1 - mass of copper, T - final equilibrium temperature.

Vapor will loose heat Q2 until it is 100°C, than vapor will turn into water, it will be condensation Q3, than it will be cool of water Q4 until it is T temperature.

Q1 = Q2 + Q3 + Q4

Q2 = 2 * m2 * (130-100), 2 - specific heat of vapor

Q3 = L * m2, L - specific heat of evaporation

Q4 = C3 * m2 * (100 - T), C3 - specific heat of water.

10*0.385*(T - 30) = 0.2*1.895*(130-100) + 0.2*2260 + 0.2*4.186*(100 - T)

3.85*T - 115.5 = 11.37 + 452 + 83.72 - 0.8372*T

T = 141.4 deg C.......But T cannot be > 130 deg C.

Assuming T > 100 deg C,

10*0.385*(T - 30) = 0.2*1.895*(130-100)

T = 32.95 which is not as we assumed.

Hence, T will be = 100 deg C.