Calorimeter Physics Problem: Steam Condensation for Temperature Increase

[smray7]

Homework Statement

50g copper calorimeter contains 250g of water @ 20 degrees C.
how much steam must be condensed into water if the final temp. of system is 50 degrees C?

Homework Equations

MwCw(T-Tw) = -MxCx(T-Tx)

The Attempt at a Solution

I converted grams to kg and celsius to kelvins.

so I'm thinking i have to add the mass and specific heat of the calorimeter to the left so now:

-MwCW(T-Tw) + McCc(T-Tc) = MsCs(T-Ts)

-(.25kg)(4189j/kg C)(373k-293k) + (.05kg)(387j/kg c)(373k-?) = Ms(2010j/kg c)(373- ?)ok so i know i need to solve for Ms but I'm not sure what my initial temps are for the copper and steam. would it be the same as the water? so 293K?

is my negative on the right side of the equation? am i right by adding McCc(T-Tc) to MwCw(T-Tw)??

 

[darkys]

Yes, you are correct to add the mass and specific heat of the calorimeter to the left side of the equation. The initial temperature of the copper and steam should be the same as the water (20°C = 293K). Also, your negative sign on the right side of the equation is correct. Now, you can solve for Ms.