i think you used the the c_p from iron (thats 0.45) (my physics book says so)
another thing: how is the composition from steel? the c_v varries for different compositions
That's exactly the problem, the question doesn't say - it just says steel. The textbook also has one table "Properties of Selected Solids at 25C" which lists the C_p (not C_v) of standard 304-grade steel at 0.46 kJ/kg*K. That's the only mention of C_p pertaining to steel I can find in the whole book, and C_v isn't for steel isn't mentioned anywhere. If I knew what kind of steel the question had in mind, I could look up the properties. If the question gave me the C_v, I could solve the question.
That's not the only issue - my heat values are way off still. The container is rigid, so no work is done, so its heatremoved = \delta(enthalphy)
Here's my current attempt:
Given:
P_1 = 1000kPa
T_1 = 200C
T_2 = 30C
0.2 kg water
12 kg steel
Rigid container - constant volume, no work
Looking in the superheated tables, I find specific volume for initial conditions is v_1 = 0.20596 m^3/kg, specific enthalpy h_1 = 2827.86 kJ/kg
The volume stays constant so I look in the saturated tables for 30C, find specific volumes for that temp
v_liq = 0.001004 m^3/kg v_gas = 32.8922 m^3/kg
Using the lever rule I find quality X_2 = 6.23 * 10^-3
In the same table, for 30C, h_liq = 125.77 kJ/kg, h_gas = 2340.48 kJ/kg
Using above quality, h_2 = 140 kJ/kg
From the first law, q = h_2 - h_1 in this case q = 2687.5 kJ/kg
I know the mass of the water is 0.2kg, so Q = 537.5 kJ
There's nothing I see wrong with this section, that Q value seems reasonable.
Problem is with steel. If I use Q = m*c*ΔT, with m as 12 and ΔT as -170, that c value of 0.46 gives me the wrong answer.
If I solve backwards using the known correct solution Q = 1353 kJ, I find the required C_v to be 0.3997 ~ 0.4.
So I'll just use that 0.4 value, claim I found it in the textbook and get the question right.
This doesn't help with entropy however.
For water, looking in the tables I got the data in the first section from, I find s_1 = 6.69 kJ/kg-K, s_2 = 0.6855 kJ/kg-K.
Thus S = m*Δs = -1.2 kJ/K
Using S = m*c*ln(T_2/T_1) for steel, I get S = -2.13 kJ/K
If the entropy of steel is positive then I get the right answer, but its negative. I'm mixed up somewhere here, just not sure where.