Chemistry Help!!! Energy and Heat?

I have two diffrent problems that I need help in my chemistry homework.








cw = 4180 J/kg.C





A 295 kg cast-iron car engine contains water as a coolant. Suppose the engine's temperature is 40 C when it is shut off and the air temperature is 13 C. The heat given off by the engine and water in it as they cool to air temperature is 4.8 X 106 J.


(Assume thespecific heat of iron is 450 J/kg .degrees C.)





What mass of water is used to cool the


engine? Answer in units of kg.


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Given:


cp;ice = 2090 J/kg. C


cp;water = 4186 J/kg. C


cp;steam = 2010 J/kg. C


Lf = 3:33 X 105 J/kg


Lv = 2:26 X 106 J/kg


How much energy is required to change a


44 g ice cube from ice at -13 C to steam at


117 C? Answer in units of J.





Thx For All Your Help =)

Chemistry Help!!! Energy and Heat?
1.


Q = m1c1/\T + m2c2/\T





m1*4186*27 + 295*450*27 = 4.8*10^6





113022*m1 + 3584250 = 4.8*10^6





m1 = (4,800,000 - 3,584,250)/113022





= 10.75 kg





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2.





Ice from - 13 to zero:





Q = 44*2090*13 = 1,195,480 J





Ice to water at zero degrees:





Q = 44*3.33*10^5 = 14,652,000 J





Water from 0 to 100 degrees:





Q = 44*4186*100 = 18,418,400 J





Water to steam at 100 degrees:





Q = 44*2.26*10^6 = 99,440,000 J





Steam from 100 to 117 degrees:





Q = 44*2010*17 = 1,503,480 J





I will leave you to add them all up!
Reply:1...Heat lost by Engine and water = 4,800kJ.


Heat lost by water = mass x 4.184kJ/kg/°C x 27°C.


= 0.112kJ x mass.


Heat lost by engine = 295kg x 0.45kJ/kg/°C x 27.


= 3,584kJ.


4,800kJ - 3,584kJ = 1,216kJ.


Mass of water = 1,216kJ / 0.112kJ = 10.9kg.





2...Ice at -13 to 0°C. (Sensible Heat).


= 44g x 2.090J/g/°C x 13°C. = 1,195.5J.


Ice at 0°C to water at 0°C (Lf).


= 44g x 334J/g = 14,696J.


Water at 0°C to 100°C. (Sensible Heat).


= 44g x 4.184 x 100°C. = 18,410J


Water at 100°C to Steam at 100°C (Lv).


= 44g x 2,260J/g = 99,440J.


Steam at 100°C to 117°C (Sensible Heat).


= 44 x 2.010J/g x 17°C. = 1,503J.





Total heat required.


= 1,503 + 99,440 + 18,410 + 14,696 + 1,195 = 135,244J.


= 135.244kJ.


....