QuestionAugust 14, 2025

A 35.0 g piece of aluminum at 405^circ C was placed into a sample of water at 23.0^circ C . The final temperature of the water and aluminum was 55.0^circ C What is the heat change for the aluminum? c_(Al)=0.921(J)/(gcdot ^circ )C q_(Al)=[?]J Round your answer to three significant figures.

A 35.0 g piece of aluminum at 405^circ C was placed into a sample of water at 23.0^circ C . The final temperature of the water and aluminum was 55.0^circ C What is the heat change for the aluminum? c_(Al)=0.921(J)/(gcdot ^circ )C q_(Al)=[?]J Round your answer to three significant figures.
A 35.0 g piece of aluminum at 405^circ C
was placed into a sample of water at
23.0^circ C . The final temperature of the
water and aluminum was 55.0^circ C
What is the heat change for the
aluminum?
c_(Al)=0.921(J)/(gcdot ^circ )C
q_(Al)=[?]J
Round your answer to three significant figures.

Solution
4.6(218 votes)

Answer

q_{Al} = -11,300 \, \text{J} (rounded to three significant figures) Explanation 1. Identify the formula for heat change Use the formula for heat change: **q = mc\Delta T**, where m is mass, c is specific heat capacity, and \Delta T is the change in temperature. 2. Calculate the change in temperature (\Delta T) \Delta T = T_{\text{final}} - T_{\text{initial}} = 55.0^{\circ}C - 405^{\circ}C = -350.0^{\circ}C 3. Substitute values into the formula q_{Al} = (35.0 \, \text{g})(0.921 \, \frac{\text{J}}{\text{g}\cdot^{\circ}C})(-350.0^{\circ}C) 4. Perform the calculation q_{Al} = 35.0 \times 0.921 \times (-350.0) = -11,284.5 \, \text{J}

Explanation

1. Identify the formula for heat change<br /> Use the formula for heat change: **$q = mc\Delta T$**, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.<br /><br />2. Calculate the change in temperature ($\Delta T$)<br /> $\Delta T = T_{\text{final}} - T_{\text{initial}} = 55.0^{\circ}C - 405^{\circ}C = -350.0^{\circ}C$<br /><br />3. Substitute values into the formula<br /> $q_{Al} = (35.0 \, \text{g})(0.921 \, \frac{\text{J}}{\text{g}\cdot^{\circ}C})(-350.0^{\circ}C)$<br /><br />4. Perform the calculation<br /> $q_{Al} = 35.0 \times 0.921 \times (-350.0) = -11,284.5 \, \text{J}$
Click to rate:

Similar Questions