The Final Equilibrium Temperature Reached?

[hatcheezy]

Homework Statement

A perfectly insulated calorimeter cup(it niether gains nor loses energy) contains 300 grams of water at 10.0 degrees Celsius. If 15.0 grams of steam at 150 degrees Celsius are added to the water, what final equilibrium temp is reached?

Homework Equations

Qlost = Qgain
(msteam)(csteam)($$\Delta$$Tsteam)+(msteam)(Lv)=(Mwater)(Cwater)($$\Delta$$Twater)

The Attempt at a Solution

(15g)(.480cal/gm$$\circ$$C)(Tf-150$$\circ$$C)+(15g)(540cal/gm)=(300g)(1.00cal/gm$$\circ$$C)(Tf-10$$\circ$$C)

(7.2)(Tf-150$$\circ$$C)+(8100)=(300)(Tf-10$$\circ$$C)

(7.2Tf)-(1080)+(8100)=(300Tf)-(3000)

(7.2Tf)+(7020)=(300Tf)-(3000)

(10020)=(292.8Tf)

(Tf)=(34$$\circ$$C)

 

[rl.bhat]

(15g)(.480cal/gmLaTeX Code: \\circ C)(Tf-150LaTeX Code: \\circ C)+(15g)(540cal/gm)=(300g)(1.00cal/gmLaTeX Code: \\circ C)(Tf-10LaTeX Code: \\circ C)
In left hand side you have substituted 0.480 csl/g/oC for th specific heat of the water and in right hand side 1.00 cal/g/oC for the same thing. How is that?