# chemistry

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100 mL of is initially at room temperature (22)C. A chilled steel rod at 2 C is placed in the water. If the final temperature of the system is 21.2 C, what is the mass of the steel bar?
Specific heat of water = 4.18J/g(C)
Specific heat of steel = 0.452J/g(C)

• chemistry -

heat lost by water = heat gained by steel

(m Cp dT) water = (m Cp dT) steel

mass steel = (m Cp dT) water / (Cp dT) steel
mass steel = (125 g) x (4.18 J/gC) x (22.0-21.1 C) / [(0.452 J/gC) x (21.1 - 2.0 C)] = 54.5 g steel

*** 2 ***
molar heat capacity is J/mole C not J/gC

4.18 J/gC x (18.0 g / mole) = 75.2 J/moleC

• chemistry -

54.5 g steel doesnt work i inputed the answer it was incorrect

• chemistry -

mass of the steel is not 54.5 g.

• chemistry -

heat lost by water + heat gained by steel = 0
massH2O x specificheatH2O x (Tf-Ti) + masssteel x specificheatsteel x (Tf-Ti) = 0
100 x 4.18 x (21.2-22) + masssteel x 0.452 x (21.2-2) = 0
Solve for mass steel. I get about 40 g but you need to solve it exactly. Watch the signs. Post your work if you get stuck. Tf is final T. Ti is initial T.

• chemistry -

• chemistry -

It would 24.9 grams
(105.5g)(0.5°C)(4.184J)=x(19.5)(0.452)

• chemistry -

The answer with 3 sig figs and units is 43.6 g

• chemistry -

38.9

• chemistry -

I pooped my pants :(

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