A volume of 115 mL of H2O is initially at room temperature (22.0 C). A chilled s

Question

A volume of 115 mL of H2O is initially at room temperature (22.0 C). A chilled steel rod at 2.00 Cis placed in the water. If the final temperature of the system is 21.5  C , what is the mass of the steel bar?

Use the following values:

specific heat of water = 4.18 J g1 C1

specific heat of steel = 0.452 J g1 C1

Express your answer to three significant figures and include the appropriate units.

Part A A volume of 115 mL of H2O is initially at room temperature (22.0 °C). A chilled steel rod at 2.00 °C is placed in the water. If the final temperature of the system is 21.5 °C, what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J g-1 °C-1 specific heat of steel = 0.452 J g-1 oC-1 Express your answer to three significant figures and include the appropriate units. View Available Hint(s) mass of the steelVale Units Submit Provide Feedback Next>

Explanation / Answer

apply the equation

H = m * Cp * T

T = Final temperature – Initial temperature

m is mass

Cp is heat capacity

H is the enthalpy

Final temperature is 21.5 C

The heat gained by the rod is equal to the heat lost by the water so

115 ml of water must be changed to grams since density of water is 1 gr/ml then it is correct to say that

115 ml are equal to 115 grams

heat lost by the water

H = 115 * 4.18 * (21.5 - 22) = -240.35 Joules

this must be equal to the heat gained by the rod

q lost = -qgained

240.35 = m * 0.452 * (21.5 - 2)

m = 240.35 / (0.452 * (21.5 - 2) = 27.269 grams

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