Possibly useful information: The specific heat of (liquid) water is cw = 4186 J/(kg C° ). The heat of fusion for ice or water is L = 3.33 x 105 J/kg The latent heat of vaporization for water is 22.57 x 105 J/kg 1 calorie = 4. 186 Joules T (Kelvin) = T (Celsius) + 273 1. A 0.040 kg ice cube at 0 Celcius is placed in an insulated box that contains 0.007kg of steam at 100 Celsius. What is the equilibrium temperature reached by the closed system. NOTE: All of the ice melts.

College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Possibly useful information:
The specific heat of (liquid) water is cw = 4186 J/(kg C° ).
The heat of fusion for ice or water is L = 3.33 x 105 J/kg
The latent heat of vaporization for water is 22.57 x 105 J/kg
1 calorie = 4.186 Joules
T (Kelvin) = T (Celsius) + 273
1. A0.040 kg ice cube at 0 Celcius is placed in an insulated box that contains 0.007kg of steam at
100 Celsius. What is the equilibrium temperature reached by the closed system. NOTE: All of
the ice melts.
Transcribed Image Text:Possibly useful information: The specific heat of (liquid) water is cw = 4186 J/(kg C° ). The heat of fusion for ice or water is L = 3.33 x 105 J/kg The latent heat of vaporization for water is 22.57 x 105 J/kg 1 calorie = 4.186 Joules T (Kelvin) = T (Celsius) + 273 1. A0.040 kg ice cube at 0 Celcius is placed in an insulated box that contains 0.007kg of steam at 100 Celsius. What is the equilibrium temperature reached by the closed system. NOTE: All of the ice melts.
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