Mass of Steam Needed for 1kg of Ice to Yield 20°C Water

AI Thread Summary
To determine the mass of steam needed to convert 1 kg of ice at 0°C to liquid water at 20°C, the latent heat of fusion (3.3 x 10^5 J/kg) is used in the calculation. The equation involves the heat required to melt the ice and raise the temperature of the resulting water. It is essential to clarify which mass variable corresponds to steam in the equation, as both sides contain the variable "m." By solving for the mass of steam, the required energy transfer can be accurately calculated. This approach ensures a clear understanding of the thermal dynamics involved in the phase change and temperature increase.
heelp
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what mass of steam at 100 c must be added to 1.ookg of ice to yeild liquid water 20c ?

this is what I have so for c*m* displacement of temperature =m *L
 
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You should know L (which I believe is 3.3 x 10^5 J/kg). Which is the Latent Heat of Fusion. You are given the mass of ice, change in temperature. Just solve for the second m in your equation. Which I'm assuming in the mass of the steam. When you have two of the same variables on each side of the equal sign, it looks confusing. Please denote "m of what".
 
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m is the mass
 
I know that m is the mass. But you have an m on each side of the equal sign. What I'm asking you to state is the "mass of what". In your equation just solve for the m that relates to steam.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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