Question

a cube of aluminum ($c_p = 0.897 \text{j/g·k}$) has a temperature of $325 \text{k}$. it is placed in a beaker of cold water. if the aluminum cools to $301 \text{k}$ as the water absorbs $815.7 \text{j}$ of heat, what is the mass of the cube?
use the formula $q = mc_pdelta t$.
$\boxed{} \text{g}$

Explanation:

Step1: Identify known values

We know that the heat lost by aluminum (\(q\)) is equal to the heat gained by water, so \(q = 815.7\space J\), \(C_p=0.897\space J/g\cdot K\), initial temperature \(T_i = 325\space K\), final temperature \(T_f=301\space K\). First, calculate the change in temperature \(\Delta T\).
\(\Delta T=T_i - T_f=325 - 301 = 24\space K\)

Step2: Rearrange the formula to solve for mass \(m\)

The formula is \(q = mC_p\Delta T\). We can rearrange it to solve for \(m\):
\(m=\frac{q}{C_p\Delta T}\)

Step3: Substitute the known values into the formula

Substitute \(q = 815.7\space J\), \(C_p = 0.897\space J/g\cdot K\) and \(\Delta T=24\space K\) into the formula:
\(m=\frac{815.7}{0.897\times24}\)
First, calculate the denominator: \(0.897\times24 = 21.528\)
Then, calculate the mass: \(m=\frac{815.7}{21.528}\approx37.9\)

Answer:

\(37.9\)