chemistry
posted by Lilit on .
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)

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 (22.021.1 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 
54.5 g steel doesnt work i inputed the answer it was incorrect

mass of the steel is not 54.5 g.
I inputed the answer it didn't work. can you please help me 
heat lost by water + heat gained by steel = 0
massH2O x specificheatH2O x (TfTi) + masssteel x specificheatsteel x (TfTi) = 0
100 x 4.18 x (21.222) + masssteel x 0.452 x (21.22) = 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. 
the answer is 42.4

It would 24.9 grams
(105.5g)(0.5°C)(4.184J)=x(19.5)(0.452) 
The answer with 3 sig figs and units is 43.6 g

38.9

I pooped my pants :(