Mastering Chemisry Part A A volume of 95.0 mL of H2O is initially at room temper

Question

Mastering Chemisry

Part A A volume of 95.0 mL of H2O is initially at room temperature (22.00 °C). A chilled steel rod at 2.00 C is placed in the water. If the final temperature of the system is 21.30 steel bar? , what is the mass of the Use the following values specific heat of water = 4.18 J/ (g·°C) specific heat of steel 0.452J/(g. °C) Express your answer to three significant figures and include the appropriate units. mass of the steelValue Units Submit Hints My Answers Give Up Review Part Part B The specific heat of water is 4.18 J/ (g·°C). Calculate the molar heat capacity of water. Express your answer to three significant figures and include the appropriate units. molar heat capacity for water = 1 Value Units Submit Hints My Answers Give Up Review Part

Explanation / Answer

Part A)

Given data:

1)For Water : Volume = 95 mL

Hence mass(mw) = 95 g…..(since density of water = 1 g/mL)

Specific heat of water (Cw) = 4.18 J/g.0C

Initial temperature (Ti) = 22 0C

Fitial temperature (Tf) = 21.30 0C

Hence, T = (Tf) -(Ti) = 21.30 – 22 = - 0.50 0C

Let us calculate heat evolved by water Qw = ?

Formual, -Q = m x C x T………(-ve sign indicate heat evolved)

-Qw = mw x Cw x T

-Qw =95 x 4.18 x (-0.50)

-Qw = -198.55 J

Qw = 198.55 J

Heat evolved by water Qw = 198.55 J.

Now

Heat gained by steel rod(Qs) = Heat evolved by water (Qw) = 198.55 J

For steel rod,

Ti = 2 0C

Tf = 21.30 0C

T = 21.30 – 2 = (Tf) -(Ti) =19.50 0C

Specific heat of steel Cs = 0.452 J/g. 0C

Mass of steel (ms) = ?

Formula,

Qs = ms x Cs x T

198.55 = ms x 0.452 x 19.50

198.55 =ms x 8.814

ms = 198.55 / 8.814

ms = 22.53 g

Mass of the steel rod = 22.53 g

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Part B

Sp. Heat of water = 4.18 J/g. 0C

Molar mass of water = 18 g

Molar heat capacity = ?

Fromula,

Molar heat capacity = sp. Heact x molar mass

Hence,

Molar heat capacity of water = sp. Heat of water x molar mass of water,

Molar heat capacity of water = 4.18 x 18

Molar heat capacity of water = 75.25 J/ 0C mole.

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