Answers to Chapter 14

1. We find the time from

                      DQ = mcDT;

                      (350W)t = (350mL)(1.00g/mL)(10^-3kg/g)(4186J/kg.C°)(50°C - 20°C), which gives t = 90s.

2. We find the temperature form 

                     heat lost = heat gained;

                     m_waterc_waterDT_water = m_glassc_glassDT_glass;

                    (135mL)(1.00g/mL)(10^-3kg/g)(4186J/kg.C°)(T - 39.2°C)

                     = (0.030kg)(840J/kg.°C )(39.2°C - 21.6°C), which gives T = 40.0°C.

3. We find the specific heat from

                    heat lost = heat gained;

                    m_xc_xDT_x = ( m_Alc_Al+ m_waterc_water+ m_glassc_glass)DT_water;

                   (0.195kg)c_x(330°C- 35.0°C)

                   = [(0.100kg)(900J/kg.C°) + (0.150kg)(4186J/kg.C°) + (0.017kg)(840j/kg.C°)](35.0C° - 12.5C°),

                  which gives  c_x = 286J/kg.C°.

4. (a) We find the heat required to reach the boiling point from

                    Q₁ = ( m_Fec_Fe+ m_waterc_water )DT

                       = [(230kg)(450J/kg.C°) + (830kg)(4186J/kg.C°)](100C° - 20°C)

                       = 2.86 x 10⁸J.

              We find the time from

                     t₁ = Q₁ /P = (2.86 x 10⁸J)/(52,000 x 10³J/h) = 5.5h.

        (b) There is no change in the temperature, so the additional heat required to change the water into steam is 

                       Q₂ =  m_waterL_steam

                            = (830kg)(22.6 x 10⁵J/kg) = 1.88 x 10⁹J.

                We find the additional time from

                        t₂ = Q2 /P = (1.88 x 10⁹J)/(52,000 x 10³J/h) = 36.1h.

                 Thus the total time required is

                       t = t₁ + t₂  = 5.5h + 36.1h = 41.6h.