4.11 Initially, an insulated rigid tank contains 3.269 kg of water at 259.7 cels

Question

4.11  Initially, an insulated rigid tank contains 3.269 kg of water at 259.7 celsius and 1 MPa.

A heater in the wall of the tank maintains the wall temperature at 600 celsius.

After a period of time, the pressure in the tank rises to 1.5 MPa.

Determine the total heat transfer (kJ) to the water in the tank.

Use specific heats from tables for water.  

Answer: 1,331.6

https://highered.mcgraw-hill.com/sites/dl/free/0073529214/395307/appdxs1_2.pdf

PLEASE SHOW ALL WORK INTERPOLATION AND

HOW TO GET ALL NUMBERS NEEDED thankyou so much

Please use these tables   and state which ones used thanks much

Explanation / Answer

moles of water = 3.269 * 1000/18 = 181.61 moles This is a constant volume process as the tank is rigid. Since water is in gaseous phase (as temperature > 100 celsius), we can use ideal gas equation PV = nRT V = nRT/P = 181.61*8.314*259.7/(1*10^6) = 0.392 m^3 again, final temperature, T_f = PV/nR = 1.5*10^(6) *0.392/(181.61*8.314) = 389.43 celsius Therefore, heat transferred = m* C_v * (delta T) = 3.269* 1.4108*(389.43 - 259.70) = 1331.6 KJ

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