Question

To make steam, you add 5.75×105J of thermal energy to 0.230 kg of water at an initial temperature of 50.0 ∘C.

Part A

Find the final temperature of the steam.

Answer 1

Total energy needed = Q = 5.75*10^5 J

m = 0.230 kg

Ti = 50 C

So,

Q = Q1 + Q2

Q1 = energy needed from 50 C water to 100 C water = m*Cw*dT1

Q2 = energy needed from 100 C water to 100 C steam = m*Lv

Q3 = energy needed from 100 C steam to T C steam = m*Cs*dT2

Cw = 4186 J/kg-C

Lv = 2.26*10^6 J/kg

Cs = 2010 J/kg-C

dT1 = 100 - 50 = 50

dT2 = T - 100

Now using these values

Q = Q1 + Q2 + Q3

Q = m*Cw*dT11 + m*Lv + m*Cs*dT2

Cs*dT2 = [Q/m - Cw*dT1 - Lv]

dT2 = [Q/m - Cw*dT1 - Lv]/Cs

T - 100 = [Q/m - Cw*dT1 - Lv]/Cs

T = 100 + [5.75*10^5/0.230 - 4186*50 - 2.26*10^6]/2010

T = 115.27 C