39. We compute with Celsius temperature, which poses no difficulty for the J/kg·K values of specific heat capacity (see Table 18-3) since a change of Kelvin temperature is numerically equal to the corresponding change on the Celsius scale. If the equilibrium temperature is T f then the energy absorbed as heat by the ice is Q I = L F m I + c w m I ( T f – 0°C), while the energy transferred as heat from the water is Q w = c w m w ( T f – T i ). The system is insulated, so Q w + Q I = 0, and we solve for T f : . () ww i FI f IC w cmT Lm T mm c − = + (a) Now T i = 90°C so 3 (4190J / kg C )(0.500kg)(90 C) (333 10 J / kg)(0.500kg)5.3 C. (0.500kg 0.500kg)(4190J / kg C )f T ⋅° ° − × == ° +⋅ ° (b) Since no ice has remained at 5.3 f TC =° , we have 0 f
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