100 grams of ice are at 0 degree C. How much energy must the ice absorb if it is
ID: 1513666 • Letter: 1
Question
100 grams of ice are at 0 degree C. How much energy must the ice absorb if it is to completely melt? If the ice were mixed with 360 grams of water initially at 30 degree C and the system is assumed to be closed, would the water be able to lose enough energy to melt all the ice? Show calculations and explain your reasoning. If instead the ice were mixed with only 100 grams of water, would all the ice melt? Show calculations and explain your reasoning. What will be the equilibrium temperature of system? Explain how you know. Consider the system and conditions described in part a. and assume a closed system. Setup an equation of the form (like in class) DeltaE_sys = 0 where you fill in the details including the latent heat term (see examples from class!). And use this equation to solve for the equilibrium temperature of the system. Remember the ice melts, the water cools to the equilibrium temperature and the melted ice warms to the equilibrium temperature as liquid water. There's a term for each in the equation and the equilibrium temperature is an unknown.Explanation / Answer
a) Energy to completely melt ice, Eice = miceLice = 100 * 333.55 = 33355 J
Maximum heat lost by 360 g water, Ew = mCT = 360 * 4.184 * (30 - 0) = 45187.2 J
Since Eice < Ew, 360 g water will be able to melt all ice.
b) Maximum heat lost by 100 g water, Ew = mCT = 100 * 4.184 * (30 - 0) = 12552 J
Since this energy is less than required energy to completely melt ice, ice will only be partially melt.
So, since all the water will reach 0 oC and ice is partially molten and at 0 oC, the quilibrium temperature of the system is 0 oC.
c) Esys = 0
=> Eice + Ew = 0
=> miceLice + (-mwCwTw) = 0
=> Tw = miceLice / mwCw = 100 * 333.55 / (360 * 4.184) = 22.14 oC
=> Equilibrium temperature, Teq = 30 - 22.14 = 7.84 oC
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