A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter containing water (the specific heat capacity of water is 4.18 J/g°C) initially at 21.1°C. The final temperature of the water is 26.8°C. Calculate the mass of water in the calorimeter.

Question

Quizplus · Accepted Answer

The heat lost by the aluminum is equal to the heat gained by the water. Using the formula $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature, we can set the heat lost by aluminum equal to the heat gained by water. For aluminum, $m = 25.0 g$, $c = 24.03 J/mol°C$, and $\Delta T = 86.4°C - 26.8°C$. For water, $c = 4.18 J/g°C$, and $\Delta T = 26.8°C - 21.1°C$. The mass of aluminum needs to be converted to moles (using aluminum's molar mass of approximately 27 g/mol) for the heat capacity calculation. However, since the specific heat capacity of aluminum is given per mole and the calculation directly involves temperatures, we can simplify the process by directly calculating the heat transfer and solving for the mass of water. The correct calculation involves equating the heat lost by aluminum to the heat gained by water and solving for the mass of water, leading to the correct answer being in the magnitude of grams, specifically 5.6 × 10^1 g, considering the significant figures and the context of the problem.

A 250 G Piece of Aluminum (Which Has a Molar Heat