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v3ra

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## Homework Statement

An aluminum container with a mass of 400 g contains 200 g of water at a temperature of 18°C. After a block of iron at a temperature of 90°C is placed in the water, the final temperature of the mixture changes to 28°C. Find the mass of the iron, assuming that there is no loss of heat to the surroundings.

## Homework Equations

mass of aluminum = 0.40 g, heat capacity of aluminum = 910 J(kg.°C)

mass of water = 0.20 g, heat capacity of water = 4200 J

mass of iron = ? heat capacity of iron = 460 J

(delta)t = change in temperature

mAcA(delta)tA +mWcW(delta)tW + mIcI(delta)tI = 0

## The Attempt at a Solution

This is the solution shown in the book:

(0.40 kg) x (910 J/(kg.°C)) x (28°C - 18°C) + (0.20 kg) x (4200 J/(kg.°C)) x (28°C - 18°C) + (mI) x (460 J/(kg.°C)) x (28°C - 90°C) = 0

3640 + 8400 - 28520 mI = 0

12040 = 28250 mI

mI = 0.422 kg

What I want to know is where they got 3640 and 8400 from, when 0.40 x 910 = 364 and 0.20 x 4200 = 840. Why did they add a zero? Is this an error in the book? I also do not know where 28 520 is from. What am I missing...

What I want to know is where they got 3640 and 8400 from, when 0.40 x 910 = 364 and 0.20 x 4200 = 840. Why did they add a zero? Is this an error in the book? I also do not know where 28 520 is from. What am I missing...