Solve Latent Heat Problem: Mass of Injected Steam

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The problem involves calculating the mass of steam injected into 1.8 kg of water to reach thermal equilibrium at 25°C. The latent heat of vaporization for water is given as 2.26e6 J/kg. The energy balance requires equating the heat gained by the water to the heat lost by the steam. The equations used include Q = cmΔT for the water and Q = Lv*ms + (4178)*ms*(100 - 25) for the steam. The discussion highlights confusion around rearranging these equations to solve for the mass of the injected steam.
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Homework Statement


Steam at 100°C is injected into 1.8 kg of water at 22°C in a well-insulated container, where it condenses and mixes with the existing water, reaching thermal equilibrium. If the final temperature of the well-mixed water is 25°C, what is the mass of the injected steam?

Latent heat of vaporization of water, Lv=2.26e6


Homework Equations


Q = cmΔT, equation for specific heat. specific heat constant c=4178
Q=Lvm, equation for energy of vaporization, m=mass.


The Attempt at a Solution


I first stated that since the the steam is losing energy during its phase change it is negative, -Qp.-Qp=Lvms
The amount of energy for the temperature change of the steam.
Qs=cmsdTs
Qs=(4178)ms(22-100)
The amount of energy for the temperature change of the water.
Qw=cmwdTw
Qw=(4178)(1.8)(25-22)

I know i need arrange together in someway but I am just a little confused on that part.
 
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Heat gained by the water Qw=(4178)(1.8)(25-22)

Heat lost by the steam Qs = Lv*ms + (4178)*ms*(100 - 25)
 
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