A 225 g of sample of aluminum was heated to 125.5 oC, then placed into 500.0 g of water at 22.5 oC ( specific heat of aluminum is 0.900 J/g oC). Calculate the final temperature of the mixture. (Assume no heat is lost to the surroundings). Express answer in 3 significant figures.

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A 225 g of sample of aluminum was heated to 125.5 oC, then placed into 500.0 g of water at 22.5 oC ( specific heat of aluminum is 0.900 J/g oC). Calculate the final temperature of the mixture. (Assume no heat is lost to the surroundings). Express answer in 3 significant figures.

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Step 1

Given information:

Mass of sample of aluminum = 225 g

Mass of water = 500.0 g

Initial temperature of a sample of aluminum = 125.5 ͦC

Initial temperature of water = 22.5 ͦC

Specific heat of aluminum = 0.900 J/g. ͦC

Step 2

The amount of heat required to raise the temperature of 1 gram of a substance by 1 ͦC is known as specific heat capacity. The expression for calculating heat capacity is represented as follows:

Step 3

Assuming no heat is lost to the surroundings and taking value of specific heat of water (4.18...

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