**Thermal Equilibrium Question**
2kg of ice is at -10°C is mixed with 15kg of water at 25°C.
What is the final temp when equilibrium is achieved.
Q = Energy
M = mass
c = Heat capacity
∆T = Change in temperature
W = Water
I = Ice
Heat capacity for water is 4200 J/Kg
Heat capacity for Ice is 2100 J/Kg
Well, i approached this question by first finding: Q Of water (lost) = Q Of Ice (Gained)
Q=M(w) x C(w) x ∆T(w) --> 15 x 4200 x (25 - T) ...
= 63000(25-T)
[T here is temperature equilibrium]
Ice would be 2 x 2100 x{i don't know what to insert here}
Note: I think the ice is also melting and requires some latent heat of fusion.
Which was [Fusion heat] = 3.34 x 10^5 J/Kg.
I'm just lost at this point from the ice ∆T.
Any suggestions what to do at that stage?
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Last edited by IshaanM8; Feb 11th 2019 at 06:20 PM.
Reason: Comma errors
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