Consider the following problem: 30 grams of water at 20 degree C is added to a g

Question

Consider the following problem: 30 grams of water at 20 degree C is added to a glass containing 50 grams of ice at 0 degree C. What will be the equilibrium temperature of the water and ice? a) Your friend says that the equilibrium temperature is -42.2 degree C. Without doing any calculations, how can you tell that this answer is incorrect? b) Below is your friend's solution. Explain what is wrong with your friend's solution. (0.05 kg)(3.33 * 105 J/kg) (0.05 kg)(4186 J/kg degree C)(T_f - 0 degree C) = (0.03 kg)(4186 J/kg degree C)(T_f - 20 degree C) T_f = -42.2 degree C c) What is the correct equilibrium temperature for this problem?

Explanation / Answer

a) It should be between 0 degree C and 20 degree C.

ice will decrease temperature of water, not increase it.

So my friend is wrong.

b) He assumed that all ice will melt, which does not.

Heat of fusion of ice = 0.05*3.33e5 which is more than maximum heat provided by water = 0.03*4186*20.

So all ice will not melt.

c) all ice will not melt, so T = 0 degree Celsius

  

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