A small block of aluminum (of mass 0.08 kg and specific heat 900 J/kg-C) at 94 C

Question

A small block of aluminum (of mass 0.08 kg and specific heat 900 J/kg-C) at 94 C is placed in a container of water (of mass 0.125 kg and specific heat 4186 J/kg-C) at 0 C. The two come to thermal equilibrium. What is the final temperature of the two bodies? calculate the heat energy of each body above T = 0 C, which is the amount of heat released to bring the body down to 0 C. Then write an expression for how much heat the combination of water and aluminum above would have at a common temperature T_F. Equate the two expressions and solve for T_F - it is mathematically doable, but is not trivial. Also, keep the variables in the equations until the very end that will make the math easier. There are no phase changes in this problem, just heat moving from the aluminum into the water. Finally, this is a tricky problem, but if you follow the hint, you should be able to solve it.

Explanation / Answer

let T be the final temperature

so apply the principle of calorimrtery

as Heat lost by Al = heat gained by water

McDT = mCdT

0.08 * 900 * (94-T) = 0.125 * 4186 * (T-0)

T = 11.37 deg C -----------<<<<<<<<<<Answer

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