Question

Exercise #11 (65 points) Specific Heat 100.0 g of ice at -15.0°C become steam at 110°C Сараcity Using the data reported in the tables, calculate the (J-g1 K1) amount of heat necessary for the process to occur H2O (s) 2.060 at standard pressure. (50 points) Draw the H2О (1), 25°C 4.184 behavior of heat versus temperature. (15 points) 2.040 H2O (g), 100°C For full credit, report on the plot all the necessary information Н2О 333.5 J/g fus H 2260 J/g ДуарH

Answer 1

The amount of heat -15⁰C ice convert to 0⁰C water

q1   = mc$\Delta$T

       = 100*2.060(0-(-15)

      = 100*2.060*15

      = 3090J

The amount of heat of fusion of ice

q2   = m$\Delta$H fus

       = 100*333.5

       = 33350J

The amount of heat water convert at 0⁰C to 100⁰C

q3   = mc$\Delta$T

        = 100*4.184*(100-0)

        = 41840J

The amount of heat steam convert 100⁰C to 110⁰C

q4 = mc$\Delta$T

      = 100*2.040*(110-100)

     = 2040J

The amount of heat of vaporization

q5 = m$\Delta$H vap

      = 100*2260

     = 226000J

The total amount of heat

q   = q1 + q2 + q3 + q4 + q5

      = 3090 + 33350 + 41840 + 2040 + 226000

     = 306320J

    = 306.32KJ >>>>answer