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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+mH+mcΔt = mcΔt+mH+mcΔt
(20g)(2.08J/g°C)(10°C)+(20g)(2268J/g)+(20g)(4.19J/g°C)(100-t) = (25g)(2.1T/g°C)(40°C)+(25g)(334J/g)+(25g)(4.19J/g°C)(t-0)
416J+45360J+83.8J/°C(100-t) = 2100J+8350J+104.75J/°C(t-0)
Combining like terms
45776J+83.8J/°C(100-t) = 10450J + 104.75J/°C(t-0)
35326J+83.8J/°C(100-t) = 104.75J/°C(t-0)
35326J+8380J/°C-83.8J/°Ct = 104.75J/°Ct
35326J+8380J/°C = 188.55J/°Ct
187.356139(1/°C)+44.444444444444444444444444444444 = t
231.8005834°C = t
232°C = t
Any help would be greatly appreciated, thanks in advance.