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kiro484
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Homework Statement
Suppose that 20.0g of steam at 110°C is mixed with 25.0g of ice at -40°C What will the final temperature be?
Ice - Specific heat capacity 2.10 J/g°C
Water - Specific heat capacity 4.19 J/g°C
Water - Latent heat of fusion 334 J/g
Water - Latent heat of vaporization 2268 J/g
Steam - Specific heat capacity 2.08 J/g°C
Homework Equations
Q=mcΔt
Q=mH
The Attempt at a Solution
Heat loss (steam) = heat gain (ice)
mcΔt+mHvap+mcΔt = mcΔt+mHfus+mcΔt
(20.0g)(2.08J/g°C)(10°C)+(20.0g)(2268J/g)+(20.0g)(4.19J/g°C)Δt = (25.0g)(2.10J/g°C)(40°C)+(25.0g)(334J/g)+(25.0g)(4.19J/g°C)Δt
416J+45360J+83.8J/°C(Δt)=2100J+8350J+104.75J/°C(Δt)
45776J+83.8J/°C(Δt)=10450J+104.75J/°C(Δt)
35326J+83.8J/°C(Δt)=104.75J/°C
[(421.5513126)(1/°C)]Δt=1.25Δt
But then I get lost and confused as to what I need to do to isolate the variable. Any help or solutions would be greatly appreciated. Thanks in advance
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