- #1
jennypear
- 16
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An insulated bucket contains 150 g of ice at 0°C. If 26.0g of steam is injected at 100°C into the bucket, what is the final equilibrium temperature of the system in °C?
this i what I've done so far
q(ice)=150g*333.5J/g=50025J
q(steam)=26g*2257J/g=58682J
q(ice) + q(ice-water) = -[q(steam) + q(steam-water)]
50025J + 4.18J/g*150g(Tf-273)=-[58682J + 4.18J/g*26g(Tf-373)]
20025J + 327J*Tf - 171171J = -58682J -108J*Tf + 40538J
-121146J + 627J*Tf = -18145 - 108J*Tf
735J*Tf=103002
Tf=140degree K
this is obviously wrong...what am i missing?
this i what I've done so far
q(ice)=150g*333.5J/g=50025J
q(steam)=26g*2257J/g=58682J
q(ice) + q(ice-water) = -[q(steam) + q(steam-water)]
50025J + 4.18J/g*150g(Tf-273)=-[58682J + 4.18J/g*26g(Tf-373)]
20025J + 327J*Tf - 171171J = -58682J -108J*Tf + 40538J
-121146J + 627J*Tf = -18145 - 108J*Tf
735J*Tf=103002
Tf=140degree K
this is obviously wrong...what am i missing?