How much heat is required to convert 500g of liquid water at 28°C into steam at 150 °C? Take the specific heat capacity of water to be 4183 J/Kg °C and the latent heat of vaporization to be 2.26 × 10^6 J/Kg.

Q = 1.404 × 10^(5) KJ

Explanation:

We are given:

Mass;m = 500 g = 0.5kg

Temperature 1;T1 = 28 °C

Temperature:T2 = 150 °C

Specific heat capacity;c_p = 4183 J/Kg °C

Latent heat of vaporization;L = 2.26 × 10^(6) J/Kg.

The heat energy needed is given by;

Q = sensible heat energy + Latent heat

Formula for sensible heat is;

Sensible heat energy = mc(t2 - t1)

Formula for Latent heat is ;

Latent heat = mL

Thus:

Q = mc(t2 - t1) + mL

Q = m[c(t2 - t1) + L]

Q = 0.5((4183(159 - 28) + (2.26 × 10^(6)))

Q = 1.404 × 10^(8) J = 1.404 × 10^(5) KJ

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

see earlier answer