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 75^oC. The cube is quickly added to 100 mL of water at 23^oC. What is the final temperature of the mixture? The specific heat of brass is 0.385 J/g^oC. The density of brass is 8.45 g/cm³.

Answer 1

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