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?

 

[nlightner]

Based on your calculations, the final equilibrium temperature reached is 34 degrees Celsius. This is the temperature at which the heat lost by the steam is equal to the heat gained by the water, resulting in thermal equilibrium. This can also be explained by the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, the energy from the steam is transferred to the water, resulting in a final temperature that is lower than the initial temperature of the steam and higher than the initial temperature of the water.