A combination of 0.250 kg of water at 25.0 Celsius and 0.400 kg of aluminum at 60.0 Celsius is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container. What is the final temperature of the water and aluminum?

Since the container is isolated, the principle of conservation of energy requires that the amount of energy Qhot lost by copper equal to the amount of energy Qcold gained by the water and aluminum:

Qcold = -Qhot

Substitute for Qcold and Qhot from Equation (*):

MwCw(Ts – Tw) + MACAI(T; – Tal) = -mCuCcu(Tf – Tcu)

mcuccu(Tcu – Tf) = MwCu(Ts – Tw) + maiCA (T; – Tai) mCuccuTcu – mCucouT; = mwew Is – mwewIw +MAICAIT– MAICAITAI Gather terms with Tg on one side:

(mwww +MAICAI + mouccu)Tg = mCucculcu +mwCwIw +MAICAITAI Solve for Tf:

Ts = mCu mCuCouTcu + mwcwTw + maiCAiTai

mwcw + malCal + mCucou Substitute numerical values:

(0.1)(387)(100) + (0.25)(4186)(20) + (0.4)(900)(26)

(0.25)(4186) + (0.4)(900) + (0.1)(387) = 23.6°C

Final answer:

Tf = 23.6°C

Explanation:

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