Answer:633.8 KJ
Explanation:
Given
mass of water[tex]\left ( m\right )=250gm[/tex]
Initial temperature[tex]\left ( T_i\right )=20^{\circ}C[/tex]
Final temperature [tex]\left ( T_f\right )=100^{\circ}C[/tex]
Specific heat of water [tex]\left ( c \right )[/tex]=4190 J/kg-k
heat of vaporization[tex]\left ( L\right )=22.6\times 10^5 J/kg[/tex]
Heat required for process[tex]\left ( Q\right )[/tex]=heat to raise water temperature from 20 to 100 +Heat to vapourize water completely
Q=mc[tex]\left ( T_f-T_i\right )+mL[/tex]
Q=[tex]0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5[/tex]
Q=[tex]\left ( 0.838+5.5\right )\times 10^5[/tex]
Q=[tex]6.338\times 10^5J=633.8 KJ[/tex]
