1) Thirty grams of silicon at 80.0°C is added to 70.0 mL of water at 20.0°C. Cal

Question

1) Thirty grams of silicon at 80.0°C is added to 70.0 mL of water at 20.0°C. Calculate the final temperature after the two substances are mixed together. The specific heat of water is 4.184 J/g°C and the specific heat of silicon is 0.7121 J/g°C.

2) A cube of brass, 2 cm on a side, is heated until the temperature of the brass is 75oC. The cube is quickly added to 100 mL of water at 23oC. What is the final temperature of the mixture? The specific heat of brass is 0.385 J/goC. The density of brass is 8.45 g/cm3.

Explanation / Answer

1) specific heat of water = 4.184 J / g C

The mass of water = 70 grams (density of water = 1g/mL)

So heat absorbed by water = Mass X specfic heat X change in temperature

Heat given by metal = Mass of metal X specfic heat of meta X change in temperature

The Heat given by meatl will be equal to heat absorbed by water, and finally they will reach to a common temperature (T2)

30 X 0.7121 X (80-T2) = 70 X 4.184 X (T2 - 20)

1708.8 - 21.36 T2 = 292.88 T2 - 5857.6

T2 = 24.07 0C

2) The density of brass = 8.45 g / cm^3

Volume of brass cube = a^3 = 2 X 2 X 2 cm^3 = 8cm^3

So mass of brass taken = Density X volume = 67.6 grams

The heat absorbed by water = Heat given by brass metal

Mass of water X specific heat of water X Change in temeprature = Mass of brass X specific heat X change in temperature

100 X 4.184 X (T2-23) = 67.6 X 0.385 ( 75-T2)

418.4 T2 - 9623.2 = 1952.25 -26.03 T2

T2 = 26.04 0C = final temperature

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