What mass of water at 25dC (d = degrees) must be allowed to come to thermal equalibrium with a 1.85kg cube of aluminum initially at 1.5*10*2dC to lower the temperature of the aluminum to 65dC? Assume the water doesn't turn to steam.
Q = Del(T)m(C), DelT= change in temperature, m = mass, C = specific heat. It doesn't actually give any equation I just assume that it deals with that one.
The Attempt at a Solution
(after conversion to Kelvin)
85K * (1.85kg) * (.91kJ/(kgK)aluminum = 143.0975kJ
143.0975kJ = 125K * m * (4.18kJ/kgK)water
I assumed this was the amount energy involved in the temperature change from aluminum at 423K to 338K, and assumed the energy is conserved since conservation of energy was a topic recently and so assumed the amount of energy that the aluminum cooled by the was amount of energy the water gained. It makes sense at first me at first except I don't understand exactly how to consider the "equilibrium" part of it, I know from calculus there's Newton's law of Cooling and so I don't know why/how that's used with varying temperatures and masses to cool something to an equilibrium instead, and neither do I get exactly how an equilibrium is being considered in this situation. Anyway, it said the answer was .845kg and I got .273kg.