Homework Help: chemistry

Posted by Lilit on Monday, October 20, 2008 at 4:09am.

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 - Sara, Monday, October 20, 2008 at 4:26am
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 - Lilit, Monday, October 20, 2008 at 5:05am
  54.5 g steel doesnt work i inputed the answer it was incorrect
chemistry - Lilit, Monday, October 20, 2008 at 4:53am
mass of the steel is not 54.5 g.
I inputed the answer it didnt work. can you please help me
chemistry - DrBob222, Monday, October 20, 2008 at 11:10am
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 - Anita, Monday, July 18, 2011 at 12:05pm
the answer is 42.4

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