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# A 100g cube of ice at 0C is dropped into 1.0kg of water thatwas originally at 80C. What is the final temperature of thewater after the ice has melted? Q=\pm mL ? or ?Qk=0 ???and ifso, how? # A 100g cube of ice at 0C is dropped into 1.0kg of water thatwas originally at 80C. What is the final temperature of thewater after the ice has melted? Q=\pm mL ? or ?Qk=0 ???and ifso, how?

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Other asked 2020-10-18
A 100g cube of ice at 0C is dropped into 1.0kg of water thatwas originally at 80C. What is the final temperature of thewater after the ice has melted?
$$\displaystyle{Q}=\pm{m}{L}?$$ or ?Qk=0 ???and ifso, how?

## Answers (1) 2020-10-19
Here in this problem there are three stages.
One is the energy required to melt ice.
Second one to rise the temperature of melted ice from0 0 C to the final temperatureT f .
And third one is the energy lost by the water to from thetemperature 80 0 C to the finaltemperature(T f ). Now according to the law of conservation of energy the amountof heat taken is equal to the amoount of heat lost.
So $$\displaystyle{m}_{{{i}{c}{e}}}{L}_{{{f}}}+{m}_{{{i}{c}{e}}}{C}_{{{w}{a}{t}{e}{r}}}{\left({T}_{{{f}}}-{0}^{{{0}}}{C}\right)}={m}_{{{w}{a}{t}{e}{r}}}{C}_{{{w}{a}{t}{e}{r}}}{\left({80}^{{{0}}}-{T}_{{{f}}}\right)}$$
Here c water = 4186J/kg. 0
C
L
$$\displaystyle{f}={3.33}\times{10}^{{{5}}}$$ J/kg And m ice = 100g = 0.1kg
mwater =1.0kg
Now substitute all the given values in the above mentionedformula and then solve for the value of the finaltemperature(Tf).

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