A 45-g piece of ice at 0 degree C is added to a sample of water at 8 degrees C. All of the ice melts and the temperature of the water decreases to 0 degree C. How many grams of water were there in the sample?

smileycellist2
2021-03-23
Answered

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tabuordg

Answered 2021-03-25
Author has **99** answers

Understand that once the ice melts, all you have is water at ${0}^{\circ}C$ . That means the water cools from ${8}^{\circ}C$ to ${0}^{\circ}C$ and the heat absorbed by the ice was enough for it to melt.

$q=\text{mass}\times \text{heat of fusion}$ the heat of fusion for water is 333J/g

q = (45 g)(333 J/g)

q = 14985 J

Now that you have the heat released by the water, use$q=mc\mathrm{\u25b3}T$

$14985=m\left(4.18\text{}\frac{J}{{g}^{\circ}}C\right)({8}^{\circ}C-{0}^{\circ}C)$

m=448.12 g mass of water

q = (45 g)(333 J/g)

q = 14985 J

Now that you have the heat released by the water, use

m=448.12 g mass of water

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