A 27.0 - g aluminum block is warmed to 65.6 °C and plunged into an insulated beaker containing 55.5 g water initially at 22.2 °C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Text Transcription for Educational Use:**

A 27.0 — g aluminum block is warmed to 65.6 °C and plunged into an insulated beaker containing 55.5 g water initially at 22.2 °C. The aluminum and the water are allowed to come to thermal equilibrium.

Assuming that no heat is lost, what is the final temperature of the water and aluminum?

The diagram depicts a simple input box with the notation:

\( T = \)  [Input Box]  °C

The interface allows for entry of the calculated final temperature, using mathematical symbols and input options provided above the input box.
Transcribed Image Text:**Text Transcription for Educational Use:** A 27.0 — g aluminum block is warmed to 65.6 °C and plunged into an insulated beaker containing 55.5 g water initially at 22.2 °C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum? The diagram depicts a simple input box with the notation: \( T = \) [Input Box] °C The interface allows for entry of the calculated final temperature, using mathematical symbols and input options provided above the input box.
Expert Solution
Step 1: Explaining the problem statement

Answer:

This problem is based on law of conservation of energy where heat lost by Al will be equal to heat gained by water because energy can't be destroyed.

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