Question

The following information is given for water at 1 atm: T) - 100.00°C Adap (100.00°C)= 2.259 x 109 J/g AH (0.00°C) - 333.5 J/g T.-0.00°C Specific heat solid - 2.100 J/g °C Specific heat liquid - 4.184 J/g °C A 32.10 g sample of solid water is initially at -23.00°C. If the sample is heated at constant pressure (P = 1 atm). kJ of heat are needed to raise the temperature of the sample to 22.00°C.

Answer 1

The amount of ice at -23^0C to 0

q1 = mc$\Delta$t

       = 32.10*2.1*(0-(-23)

       = 32.10*2.1*23

         = 1550.43J

The mount of heat of fusion

q2   = m$\Delta$H fusion

       = 32.1*333.5    = 10705.35J

The amount of heat 0^0C to 22^0C

q3   = mc$\Delta$T

      = 32.10*4.184*(22-0)    = 2954.7408J

The total amount of heat

q   = q1 + q2 + q3

      = 1550.43 + 10705.35 + 2954.7408

       = 15210.5J

      = 15.2105KJ >>>>>answer