Question

A  1  kg  steamActivity 2 0 water  mixture  at  1.0  MPa  is  contained  in  an  inflexible  tank.  Heat  is  added  until  the  pressure  rises to 3.5 MPa  and 400 C. Determine  the heat added

Accepted Answer

M = 1 KG\ P_1 = 1.0\ MPA\ P_2 = 3.5\ MPA\ U_1 = 2401.41\FRAC{ KJ}
{KG}\ U_2= 2926.4 \FRAC{ KJ} {KG}\ T = 400°C
HEAT TRANSFERRED = M(U_2-U_1) =1(2926.4-2401.41) = 524.99= 524.99 KJ

Question #221984

Expert's answer

$m = 1 kg\\ P_1 = 1.0\ MPa\\ P_2 = 3.5\ MPa\\ U_1 = 2401.41\frac{ kJ} {kg}\\ U_2= 2926.4 \frac{ kJ} {kg}\\ T = 400°C$
Heat transferred = $m(U_2-U_1) =$ $1(2926.4-2401.41) = 524.99$ $= 524.99 kJ$