please give me the correct answer and show work Name Maula l

   please give me the correct answer and show work.

Name Maula l\'adaan PHYS 2211 Show all work, including units! Any formula or number required will be found on the ast page. As always, we\'ll ignore air resistance. 1) An electric heater and 0.022 kg of (cce 2050 Jkg ec, Lt 333,000 Jkg, cwater 4186 JMkg C) at -10°C are inside a thermally insulated container. The heater\'s output is 575 W. Assume that we always have thermodynamic equilibrium there are places in the container hotter or colder than anywhere else). a) Ater 10 seconds of operation, what is the temperature inside the container and how much b) After 20 seDands of operation, what is the temperature inside the container and how much ice is left?

Solution

specific heat of ice C = 2050 J/kg-C

pwoer rating of the heater = 575 W

energy supplied by he heater in 10s    Q = 575*10 = 5750 J

temperature of ice = -10 C

heat required to raise the temperature to 0 ⁰C   = 2050*0.022*10 = 451 J

heat required to melt the ice = 333000*0.022 = 7326 J

The heat supplied is not sufficient to melt the complete ice

heat available to melt the ice after reaching 0 C = 5750-451 = 5299 J

amount of ice that could melt = 5299/7326 = 0.7233

72.33 % of the ice would melt after 10s and the temperature remains at 0 C

amout of ice left = 0.022*0.2767 = 0.0061 kg

b) After 20 sec 5750 J of more heat will be available.

            Total heat energy supplied = 5750*2 = 11500 J

   heat to raise ice temp to 0 C    = 451 J

heat to melt the ice                      = 7326 J

               Total                              = 7778 J

heat available to increase the water temp     = 11500 - 7778 = 3722 J

specific heat of water Cw = 4186J/kg-C

final temperature of water = 3722/(4186*0.022) = 40.42 ⁰C

all the ice completly melts after 20s